FROM  -THE-  LI  BRARY-  OF 
•WILL1AM-A  HILLEBRAND 


PROBLEMS 

IN 

ALTERNATING  CURRENT  MACHINERY 


McGraw-Hill  BookCompariy 

PwSGsfiers 


Ele  c  trie  al  Wor  Id         The  Engineering  and  Mining  Journal 
En5ineering  Record  Engineering  News 

Railway  Age  Gazette  American  Machinist 

Signal  E,ngin<?9r  American  Engineer 

Electric  Railway  Journal  Coal  Age 

Metallurgical  and  Chemical  Engineering  P  o  we  r 


PROBLEMS  IN  ALTERNATING 
CURRENT  MACHINERY 


BY 
WALDO  V.  LYON 


INSTRUCTOR   IN    ELECTRICAL   ENGINEERING,    MASSACHUSETTS 
INSTITUTE   OF   TECHNOLOGY. 


FIRST  EDITION 


McGRAW-HILL  BOOK  COMPANY,  INC. 
239  WEST  39TH  STREET,  NEW  YORK 

6  BOUVERIE  STREET,  LONDON,  E.  C. 
1914 


COPYRIGHT,  1914,  BY  THE 
MCGRAW-HILL  BOOK  COMPANY,  INC. 


THE. MAPLE- PRESS. YORK- PA 


PREFACE 

These  problems  chiefly  concern  the  theory  of  the  operation  of 
alternating  current  machinery,  and  are  such  as  we  give  to  the 
fourth  year  students  in  electrical  engineering  in  this  subject. 

In  each  chapter  the  problems  of  a  similar  nature  are  grouped 
together,  and  those  of  each  group  are  then  arranged  in  the 
approximate  order  of  their  difficulty.  The  groups  in  each  chapter 
follow  each  other  in  as  logical  an  order  as  possible,  both  from  point 
of  difficulty  and  the  presentation  of  the  subject. 

In  order  that  this  collection  of  problems  may  be  useful  among 
different  classes  of  students  a  large  variety  has  been  included, 
ranging  from  the  very  simple  to  those  of  considerable  difficulty. 
Wherever  it  is  essential  the  data  have  been  taken  from  actual 
apparatus.  This  was  possible  through  the  courtesy  of  two  of 
the  large  manufacturing  companies.  In  some  of  the  problems 
so  few  data  are  given  that  approximate  methods  of  solution 
must  be  used,  but  care  has  been  taken  to  so  state  them  that 
the  errors  thus  introduced  need  not  be  large.  This  lack  of  data 
is  frequently  met  in  practice. 

It  has  been  thought  best  not  to  give  introductory  paragraphs 
for  each  chapter  as  was  done  in  the  preceding  volume  of  prob- 
lems inasmuch  as  they  would  have  to  be  of  considerable  length 
to  be  of  much  value. 

The  answers  to  the  problems  will  probably  be  ready  for 
publication  in  the  fall  of  1914.  They  will  be  available  to  all 
but  undergraduate  students  at  a  nominal  price.  Undergraduate 
students  can  obtain  them  only  on  the  recommendation  of  their 
instructors. 

WALDO  V.  LYON. 

MASSACHUSETTS  INSTITUTE  OF  TECHNOLOGY, 
December,  1913. 


995890 


CONTENTS 

CHAPTER  PAGE 

I.  TRANSFORMERS  (98  PROBLEMS) 1 

Iron  loss,  magnetizing  current,  number  of  turns,  effective  resist- 
ance and  reactance,  copper  loss,  regulation,  efficiency,  auto- 
transformer,  induction  regulator,  parallel  operation. 

II.  SYNCHRONOUS  GENERATORS  (99  PROBLEMS) 24 

Generated  electromotive  force,  form  factor,  harmonics,  concen- 
trated and  distributed  windings,  leakage  reactance,  armature 
reaction,  synchronous  reactance  voltage,  regulation-comparison 
of  methods,  effect  of  power  factor,  efficiency,  parallel  operation, 
hunting. 

III.  SYNCHRONOUS  MOTORS  (50  PROBLEMS) 52 

Excitation,  efficiency,  power  factor,  maximum  current,  break- 
down, effect  of  added  resistance  and  reactance,  transmission  line 
regulation,  power  factor  compensation. 

IV.  INDUCTION  MOTORS  (80  PROBLEMS) 68 

Flux  distribution,  exciting  current,  iron  loss,  winding  pitch, 
starting  current,  torque,  speed  regulation,  breakdown  torque, 
effect  of  added  resistance  and  reactance,  Heyland  diagram, 
calculation  of  characteristics,  concatenation,  induction  gen- 
erator, necessary  synchronous  apparatus,  calculation  of  char- 
acteristics. 

V.  ROTARY  CONVERTERS  (61  PROBLEMS) 93 

Voltage  ratio,  effect  of  flux  distribution,  relative  outputs,  trans- 
former capacities,  armature  reaction,  excitation,  series  field 
turns,  efficiency. 

VI.  POLYPHASE  CIRCUITS  (75  PROBLEMS) 105 

Delta  and  Y  connection,  balanced  and  unbalanced  loads, 
neutral  current,  power  measurement  with  unbalanced  loads, 
transformer  connections  and  relative  capacities,  auto- trans- 
formers, transmission  line  regulation  and  efficiency,  combined 
efficiency  and  regulation  of  generator,  line  and  transformers, 
induction  and  synchronous  motor  loads. 

VII.  NON-SINUSOIDAL  WAVES  (44  PROBLEMS) 126 

Phase  and  line  voltages,  two-phase  to  three-phase  transforma- 
tion, neutral  current,  power  measurement,  effect  of  inductance 
and  capacity,  cyclic  order  of  voltages,  unbalanced  loads,  trans- 
former connections. 


Vll 


INTRODUCTION 

The  great  importance  of  problem  work  in  the  training  of  stu- 
dents of  engineering  is  now  generally  recognized.  No  other  work 
so  efficiently  develops  analytical  power  and  clear,  logical  think- 
ing, so  necessary  to  success  in  the  engineering  profession.  Yet 
notwithstanding  the  importance  of  problem  work  in  general,  the 
first  consistently  developed  book  of  electrical  engineering  prob- 
lems was  that  prepared  by  Mr.  Lyon  in  1908,  its  wide  use  being 
conclusive  evidence  of  the  needs  that  were  felt  among  both 
teachers  and  students  for  such  a  work,  and  of  the  appreciation  of 
the  importance  of  the  training  which  it  exemplifies. 

It  requires  a  special  gift  to  originate  and  develop  problems 
which  shall  give  sound  training  in  the  fundamentals  of  engineering 
and  whose  solution  shall  not  only  interest  the  student,  but  develop 
his  intellectual  power  as  well.  In  general  the  problems  must  be 
closely  related  to  engineering  practice,  graded  as  to  difficulty, 
and  must  carefully  avoid  being  mere  mathematical  puzzles.  Mr. 
Lyon,  as  his  earlier  book  also  shows,  has  a  most  unusual  ability  in 
the  preparation  of  problems  for  the  electrical  engineering  field. 

The  present  collection  of  problems  relating  to  electrical 
machinery,  more  particularly  in  the  field  of  alternating  current 
engineering,  has  been  prepared  with  the  same  point  of  view  as 
was  Mr.  Lyon's  earlier  work,  and  should  likewise  prove  most 
useful  to  both  instructor  and  student. 

H.  E.  CLIFFORD. 

HARVARD  UNIVERSITY, 
December,  1913. 


IX 


PROBLEMS  IN  ALTERNATING 
CURRENT  MACHINERY 


CHAPTER  I 
TRANSFORMERS 


1.  The  iron  loss  in  a  reactor  is  240  watts  of  which  48  watts  is 
due  to  eddy  currents.     If  the  amount  of  iron  in  the  magnetic 
circuit  were  doubled  by  doubling  the  cross-section  of  the  core, 
what  would  be  the  iron  loss  for  the  same  impressed  voltage  and 
frequency?     Neglect  the  resistance  drop. 

2.  The  iron  loss  in  a  reactor  is  312  watts  of  which  86  watts  is 
due  to  eddy  currents.     If  a  similar  reactor  were  constructed  in 
which  laminations  of  twice  the  thickness  were  used,  what  would 
be  the  iron  loss  for  the  same  impressed  voltage  and  frequency? 
Neglect  the  resistance  drop. 

3.  When  110  volts  at  30  cycles  is  impressed  on  a  reactor  the 
iron  loss  is  276  watts  of  which  204  watts  is  due  to  hysteresis.     If 
the  impressed  voltage  and  frequency  are  both  doubled  what  will 
be  the  iron  loss?     Neglect  the  resistance  drop. 

4.  When  110  volts  at  60  cycles  is  impressed  on  a  reactor,  the 
iron  loss  is  248  watts  of  which  25  per  cent,  is  due  to  eddy  currents, 
(a)  What  will  be  the  iron  loss  when  220  volts  at  60  cycles  is 
impressed  on  the  reactor?     (b)  What  will  be  the  iron  loss  when 
110  volts  at  30  cycles  is  impressed  on  the  reactor?     Neglect  the 
resistance  drop. 

5.  A  reactor  has  two  electric  circuits  having  the  same  number 
of  turns  which  may  be  connected  in  series  or  in  parallel.     When 
they  are  connected  in  series  across  220-volt,  60-cycle  mains  the 
iron  loss  is  326  watts,  of  which  89  watts  is  due  to  eddy  currents. 
What  will  be  the  iron  loss  if  the  coils  are  connected  in  parallel 
across  the  same  mains?     Neglect  the  resistance  drop. 

6.  With  425  volts  at  25  cycles  impressed  on  the  low-tension 
winding  of  a  transformer  2500  watts  is  supplied  at  no  load.     If 
the  frequency  of  this  impressed  voltage  is  increased  to  40  cycles 

1 


2        PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

without  changing  its  root-mean-square  value  or  form  factor,  2020 
watts  is  supplied  at  no  load.  What  is  the  division  between  the 
eddy  current  and  hysteresis  losses  at  25  cycles? 

7.  With  440  volts  at  50   cycles  impressed  on  the  low-tension 
winding  of  a  transformer  641  watts  is  supplied  at  no  load.     If 
the  frequency  and  voltage  are  each  reduced  50  per  cent,  without 
changing  the   form   factor  278   watts   is  supplied  at  no  load. 
What  is  the  division  between  the-eddy  current  and  hysteresis 
losses  at  50  cycles? 

8.  With  440  volts  at  60  cycles  impressed  on  the  low-tension 
winding  of  a  transformer  371  watts  is  supplied  at  no  load.     If  the 
voltage  is  reduced' 50  per  cent,  without  changing  the  frequency  or 
the  form  factor  114  watts  is  supplied  at  no  load.     What  is  the  di- 
vision between  the  eddy  current  and  hysteresis  losses  at  440  volts? 

9.  If  a  simple  harmonic  electromotive  force  of  2200  volts  at 
60  cycles  is  impressed  on  the  low-tension  winding  of  a  transformer 
the  core  loss  is  940  watts,  23  per  cent,  of  which  is  due  to  eddy 
currents,     (a)  What  will  be  the  core  loss  if  an  electromotive 
force  of  the  same  effective  value  and  frequency  but  with  a  form 
factor  of  1.051  is  impressed  on  the  low-tension  winding?     (b) 
What  will  be  the  core  loss  if  an  electromotive  force  of  the  same 
effective  value  and  frequency  but  with  a  form  factor  of  1.22  is 
impressed  on  the  low-tension  winding? 

10.  The  magnetic  circuit  of  a  transformer  has  a  mean  length 
of  77.3  in.  and  an  average  cross-section  of  34.4  sq.  in.      The 
low-tension  winding  has  399  turns.     Find  the  core  loss,  the  no- 
load  current,  and  the  power  factor  at  no  load  when  2300  volts  at 
60  cycles  is  impressed  on  the  low-tension  winding.     The  curve  of 
core  loss  at  60  cycles  and  flux  density,  and  the  B-H  curve  are 
given  by  the  data  on  the  following  page : 

11.  A  magnetic   circuit  has  a  mean  length  of  69.5  in.  and  an 
average  cross-section  of  30.0  sq.  in.     It  is  wound  with  a  coil  of 
94  turns.     A  constant  voltage  of  440  volts  at  50  cycles  is   im- 
pressed on  this  coil.     How  long  an  air-gap  would  it  be  necessary 
to  cut  in  the  magnetic  circuit  in  order  that  the  coil  would  take  a 
current  20  times  as  great?     What  is  the  power  factor  before  and 
after  cutting  the  air-gap?     Neglect  the  resistance  of  the  coil  and 
use  the  magnetic  data  given  in  problem  10.     Assume  that  the  core 
loss  at  50  cycles  is  80  per  cent,  of  the  loss  at  60  cycles. 

1  Flat  topped. 

2  Peaked. 


TRANSFORMERS 


Ordinates 

Abscissae 

Flux  density 
Kiloiines  per  sq.  in. 

Magnetizing  force 
„    4*-  NI 

H~w  l^- 

Core  loss  at  60  cycles 
Watts  per  cu.  in. 

40.0 

2.0 

0.132 

53.0 

3.0 

0.206 

61.5 

4.0 

0.262 

67.0 

5.0 

0.300 

73.6 

7.0 

0.352 

80.0 

10.0 

0.406 

86.0 

15.0 

0.457 

89.6 

20.0 

0.490 

This  data  is  for  a  good  grade  of  sheet  steel. 

For  all  of  the  problems  to  which  this  data  applies  assume  that  at  the 
working  flux  density  the  joints  in  the  magnetic  circuit  require  75  additional 
ampere  turns. 

12.  The  magnetic  circuit  of  a  5000-kv.-a.   transformer  has  a 
mean  length  of  140  in.  and  an  average  cross-section,  of  613  sq. 
in.     The  number  of  turns  in  the  high-tension  winding  is  such  that 
with  the  rated  voltage  of  52,000  volts  at  50  cycles  impressed  on 
this  winding,  17,500  watts  is  supplied  at  no  load.     How  many 
turns  has  the  high-tension  winding?     What  are  the  no-load  cur- 
rent and  power  factor?     Use  the  magnetic  data  given  in  problem 
10.     Assume  that  the  core  loss  at  50  cycles  is  80  per  cent,  of  the 
loss  at  60  cycles. 

13.  If  the  transformer  described  in  problem  12  is  connected  to 
a  66,000-volt,  50-cycle  circuit,  what  will  be  the  no-load  current 
and  power? 

14.  The  magnetic  circuit  of  a  lOOO-kv.-a.   transformer  has  a 
mean  length  of  95  in.  and  an  average  cross-section  of  278  sq.  in. 
This  transformer  is  designed  to   operate  from   a  66,000-volt, 
60-cycle  circuit.     If  the  maximum  flux  density  is  to  be  70,000 
lines  per  square  inch  how  many  turns  should  the  high-tension 
winding  have?     What  are  the  no-load  current  and  power?     Use 
the  magnetic  data  given  in  problem  10. 

15. 1  At  no  load  a  500-kv.-a.  transformer  takes   a  current  of 

1  Assume  that  the  original  flux  density  is  70  kilolines  per  sq.  in.,  and  that 
the  relation  between  the  flux  density  and  the  permeability  for  this  quality 
of  iron  is  expressed  by: 

5=102-0.017^ 

between  B  equals  45  kilolines  and  B  equals  95  kilolines  per  sq.  in.  In  prob- 
lems where  this  applies  the  effect  of  the  joints  in  the  magnetic  circuit  will 
be  neglected. 


4        PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

3.35  amperes  and  a  power  of  2960  watts,  when  the  voltage 
impressed  on  the  high-tension  side  is  11,000  volts  at  60  cycles. 
What  current  and  power  will  this  transformer  take  at  no  load  if 
12,700  volts  at  60  cycles  is  impressed  on  the  high-tension  winding? 
Assume  that  the  core  loss  varies  as  the  1.7  power  of  the  flux 
density. 

16. '  A  50-kv.-a.  transformer  takes  a  current  of  0.1  ampere  and 
a  power  of  641  watts  at  no  load  when  30,000  volts  at  50  cycles  is 
impressed  on  the  high-tension  winding.  What  current  and  power 
will  this  transformer  take  at  no  load  if  30,000  volts  at  60  cycles  is 
impressed  on  the  high-tension  winding?  Assume  that  the  core 
loss  varies  as  the  1.3  power  of  the  frequency  and  the  1.7  power  of 
the  flux  density. 

17.  The  resistance  to  direct  current  of  a  reactor  which  has  a 
laminated  magnetic  circuit  is  0.2  ohm.  When  110  volts  at  60 
cycles  is  impressed  on  this  reactor  the  current  is  10  amperes  and 
the  power  is  550  watts.  What  are  the  apparent  resistance  and 
reactance  of  the  reactor?  What  is  the  actual  inductance  of  the 
reactor  at  this  voltage  and  frequency? 

18. l  When  110  volts  at  60  cycles  is  impressed  on  a  reactor 
which  has  a  laminated  magnetic  circuit  it  absorbs  500  watts.  At 
this  time  the  inductance  of  the  reactor  is  25  mil-henrys  and  the 
resistance  is  negligibly  small,  (a)  What  are  the  current  and 
power  factor?  What  are  the  apparent  resistance  and  reactance 
of  the  reactor?  (b)  If  146  volts  at  60  cycles  is  impressed  on 
this  reactor  what  will  be  the  current  and  power  factor?  What 
will  be  the  apparent  resistance  and  reactance  of  the  reactor? 
Assume  that  the  iron  loss  varies  as  the  1.7  power  of  the  flux 
density. 

19.  A  reactor  with  a  laminated  magnetic  circuit  has  a  negligibly 
small  resistance  to  direct  current.  When  110  volts  at  25  cycles 
is  impressed  on  this  reactor  the  current  is  20  amperes  and  the 
power  is  1000  watts.  If  a  very  small  air-gap  is  cut  in  the  mag- 

1  Assume  that  the  orignal  flux  density  is  70  kilolines  per  sq.  in.,  and  that 
the  relation  between  the  flux  density  and  the  permeability  for  this  quality 
of  iron  is  expressed  by : 

5=102-0.017^ 

between  B  equals  45  kilolines  and  B  equals  95  kilolines  per  sq.  in.  In  prob- 
lems where  this  applies  the  effect  of  the  joints  in  the  magnetic  circuit  will 
be  neglected. 


TRANSFORMERS  5 

netic  circuit  the  current  becomes  100  amperes.  By  what  amount 
are  the  apparent  resistance  and  reactance  of  the  reactor  changed 
by  the  introduction  of  the  air-gap? 

20.  A  reactor  with  a  laminated  magnetic  circuit  has  a  resistance 
to  direct  current  of  0.5  ohm.     When  110  volts  at  60  cycles  is 
impressed  on  this  reactor  the  apparent  resistance  and  reactance 
are  2 . 0  ohms  and  4.0  ohms  respectively.     What  is  the  actual  in- 
ductance of  the  reactor  at  this  voltage? 

21.  At  no  load  with  220  volts  at  60  cycles  impressed  on  a 
transformer  the  ratio  of  eddy  current  loss  to  hysteresis  loss  is 
1:3.     What  is  this  ratio  if  220  volts  at  50  cycles  is  impressed  on 
the  transformer? 

22.  At  no  load  with  220  volts  at  60  cycles  impressed  on  a 
transformer  the  ratio  of  eddy  current  loss  to  hysteresis  loss  is  1:3. 
What  is  this  ratio  if  110  volts  at  60  cycles  is  impressed  on  the 
transformer? 

23.  At  no  load  with  110  volts  at  60  cycles  impressed  on  a  trans- 
former the  current  is  1.3  amperes  and  the  power  is  52  watts. 
What  will  be  the  no-load  current  and  power  taken  by  a  trans- 
former similar  to  this  in  every  particular,  except  that  the  cross- 
section  of  the  magnetic  circuit  is  double,  when  220  volts  at  60 
cycles  is  impressed  on  it? 

24.  At  no  load  with  110  volts  at  30  cycles  impressed  on  a  trans- 
former the  current  is  1.2  amperes  and  the  power  is  53  watts,  of 
.which  14  watts  is  dissipated  in  eddy  currents.     What  will  be  the 

current  and  power  at  no  load  if  220  volts  at  60  cycles  is  impressed 
on  this  transformer? 

25.  The  high-tension  winding  of  a  transformer  consists  of  two 
coils  which  may  be  connected  in  series  or  in  parallel.     When 
these  coils  are  connected  in  series  across  2200  volts  at  60  cycles 
the  current  is  0.30  amperes  and  the  power  is  140  watts  on  open 
circuit.     What  is  the  no-load  current  and  power  if  the  coils  are 
connected  in  parallel  across  1100  volts  at  60  cycles? 

26. 1  In  problem  25  what  is  the  no-load  current  and  power  when 
the  coils  are  connected  in  parallel  across  1550  volts  at  60  cycles? 

1  Assume  that  the  original  flux  density  is  70  kilolines  per  sq.  in.,  and  that 
the  relation  between  the  flux  density  and  the  permeability  for  this  quality 
of  iron  is  expressed  by: 

5=102-0.017^ 

between  B  equals  45  kilolines  and  B  equals  95  kilolines  per  sq.  in.  In  prob- 
lems where  this  applies  the  effect  of  the  joints  in  the  magnetic  circuit  will 
be  neglected. 


6        PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

Assume  that  the  core  losses  vary  as  the  1.7  power  of  the  flux 
density. 

27.  The  high-tension  winding  of  a  transformer  consists   of 
two  coils  which  may  be  connected  in  series  or  in  parallel.     When 
these  coils  are  connected  in  parallel  across  22,000  volts  at  60 
cycles  the  no-load  current  is  0.077  amperes  and  the  power  is  371 
watts,  28  per  cent,  of  which  is  eddy  current  loss.     What  are  the 
no-load  current  and  power  when  the  coils  are  connected  in  series 
across  22,000  volts  at  30  cycles? 

28.  At  no  load,  with  460  volts  impressed  on  the  primary  of  a 
transformer,  the  current  is  0.94  ampere  and  the  power  is  122 
watts.     Another  transformer  is  similar  to  this  one  except  that 
the  primary  winding  has  twice  as  many  turns  and  the  cross-sec- 
tion of  the  magnetic  circuit  is  one-half  as  great.     What  no-load 
current  and  power  will  this  transformer  take  with  460  volts  im- 
pressed on  its  primary  winding? 

29.  In  a  transformer  having  a  ratio  of  transformation  of  NI:  N2 
the  windings  are  so  designed  that  the  current  densities  in  the 
primary  and  secondary  are  the  same.     What  is  the  ratio  of  pri- 
mary to  secondary  resistance  if  both  windings  have  the  same 
mean  length  per  turn?     What  is  this  ratio  if  the  mean  length 
per  turn  of  the  primary  is  15  per  cent,  greater  than  that  of  the 
secondary?     (b)  If  the  current  density  in  the  primary  had  been 
15  per  cent,  greater  than  in  the  secondary  what  would  have 
been  this  ratio  of  the  resistances  if  the  windings  had  the  same 
mean  length  per  turn? 

30.  In  a  transformer  having  a  ratio  of  transformation  of  NI  :  N2 
the  windings  are  so  designed  that  the  heating  losses  in  the  pri- 
mary and  secondary  are  the  same.     What  is  the  ratio  of  primary 
to  secondary  resistance  if  both  windings  have  the  same  mean 
length  per  turn?     What  is  this  ratio  if  the  mean  length  per  turn 
of  the  primary  winding  is  15  per  cent,  greater  than  that  of  the 
secondary?     (b)  If  the  heating  loss  in  the  primary  had  been  15 
per  cent,  greater  than  in  the  secondary  what  would  have  been 
this  ratio  if  both  windings  had  the  same  mean  length  per  turn? 

31.  What  effect  will  be  produced  on  the  iron  losses  of  a  trans- 
former and  on  the  kilowatt  rating,  on  the  basis  of  the  same  total 
losses,  by  doubling  the  number  of  turns  in  the  primary  and  secon- 
dary windings  and  halving  the  cross-section  of  the  wires:  first, 
when  the  impressed  voltage  is  doubled;  and  second,  when  the 
impressed  voltage  is  unchanged  and  the  cross-section  of  the  mag- 


TRANSFORMERS  7 

netic  circuit  is  halved?     Assume  that  originally  the  iron  and 
copper  losses  are  equal. 

32.  In  a  given  transformer  the  ratio  of  the  iron  losses  to  the 
copper  losses  at  full  load  is  0.7.     Another  transformer  with  the 
same  rating  has  a  magnetic  circuit  of  the  same  length  but  of  25 
per  cent,  greater  cross-section,  due  to  which  the  mean  length  of 
one  turn  of  the  winding  is  10  per  cent,  greater.     Both  transformers 
are  wound  with  the  same  size  wire.     The  second  transformer 
is  designed  so  as  to  have  the  same  maximum  flux  density  as  the 
first.     What  is  the  ratio  of  the  total  losses  at  full  load,  and  what 
is  the  ratio  of  the  iron  losses  to  the  copper  losses  in  the  second 
transformer  at  full  load?     Compare  the  amounts  of  copper  in 
the  two  transformers.     Both  transformers  are  designed  for  the 
same  voltage. 

33.  Two  transformers  have  identical  magnetic  cores  and  the 
same  rated  output.     Both  are  wound  for  the  same  voltage  but  in 
the  first  the  iron  losses  are  25  per  cent,  greater  than  in  the  second 
due  to  a  higher  flux  density.     In  the  first  transformer  the  ratio  of 
the  iron  losses  to  the  copper  losses  at  full  load  is  0.7.     Both 
transformers  have  the  same  size  of  wire  in  their  windings,  but  the 
mean  length  of  one  turn  is  5  per  cent,  greater  in  the  second  than  in 
the  first.     What  is  the  ratio  of  the  iron  losses  to  the  copper  losses 
at  full  load  in  the  second  transformer?     What  is  the  ratio  of  the 
total  losses  at  full  load  in  the  two  transformers?     Compare  the 
amounts  of  copper  in  the  two  transformers.     Assume  that  the  iron 
losses  vary  as  the  1.7  power  of  the  flux  density. 

34.  Two  transformers  have  identical  magnetic  cores  and  the 
same  rated  output.     Both  are  wound  for  the  same  voltage  but  the 
first  has  15  per  cent,  more  turns  in  all  of  its  coils  than  the  second, 
due  to  which  the  mean  length  of  one  turn  is  5  per  cent,  greater  in 
the  first.  The  windings  of  each  are  of  the  same  size  wire.     Assume 
that  the  iron  losses  vary  as  the  1.7  power  of  the  flux  density. 
The  ratio  of  the  iron  losses  to  the  copper  losses  at  full  load  is  0.7 
in  the  first  transformer.     What  is  the  ratio  of  the  iron  losses  to 
the  copper  losses  at  full  load  in  the  second  transformer?     What  is 
the  ratio  of  the  total  losses  at  full  load  in  the  two  transformers? 

35.  In  a  given  transformer  the  ratio  of  the  iron  losses  to  the 
copper  losses  at  full  load  is  0.75.     Another  transformer  with  the 
same  radiating  capacity  has  a  magnetic  core  of  the  same  length 
but  with  a  cross-section  15  per  cent,  larger.     The  winding  of  the 
second  transformer  is  designed  so  that  the  iron  losses  are  one- 


8        PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

third  greater  than  the  iron  losses  of  the  first.  The  same  size  of 
wire  is  used  in  winding  both  transformers  but  the  mean  length 
of  one  turn  is  5  per  cent,  greater  in  the  second  on  account  of  the 
larger  magnetic  core.  Assume  that  the  iron  losses  vary  as  the 
1.7  power  of  the  flux  density.  Both  transformers  have  the  same 
voltage  rating.  The  first  transformer  has  a  full-load  capacity  of 
100  kw.  and  a  full-load  efficiency  of  97.9  per  cent,  at  unit  power 
factor.  On  the  basis  of  the  same  total  losses  what  should  be  the 
full-load  rating  of  the  second  transformer?  What  is  the  ratio 
of  the  iron  losses  to  the  copper  losses  at  full  load  in  the  second 
transformer? 

36.  Three    equal  10  to  1    transformers,  each  of  100  kv.-a. 
capacity,  are  arranged  with  both  the  primaries  and  secondaries 
in  Y  to  receive  power  from  a  3-phase,  2200-volt  circuit.     The 
losses  in  each  transformer  at  full  load  are  1960  watts  of  which  940 
are  core  losses.     If  these  transformers  are  connected  in  delta  to 
this  circuit,  what  power  will  they  deliver  without  exceeding  their 
full-load  heating  losses?     Assume  that  the  iron  losses  vary  as  1.7 
power  of  the  flux  density. 

37.  The    name    plate  of  a    transformer   gives  the    following 
data:  100  kv.-a.,  6600  :  220  volts,  25  cycles.     The  percentage 
distribution  of  the  losses  at  full  load  is:  copper  loss,  1.0  per  cent. ; 
eddy  current  loss,  0.3  per  cent.;  hysteresis  loss  0.8  per  cent.     If 
the  insulation  of  the  transformer  will  safely  stand  double  volt- 
age what  should  be  its  rating  on  the  basis  of  the  same  total 
heating  losses  when  taking  power  from  an  11,000-volt,  50-cycle 
circuit? 

38.  The  name  plate  of  a  transformer  gives  the  following  data: 
500  kv.-a.,  13,200  :  400  volts,  50  cycles.     At  full  load  the  copper 
loss  is  4600  watts  and  the  iron  loss  is  2800  watts.     Of  the  latter  22 
per  cent,  is  due  to  eddy  currents.     If  this  transformer  is  to  receive 
power  from  an  11,000-volt,  60-cycle  circuit  what  should  be  its 
full-load  rating  on  the  basis  of  the  same  total  heating  losses? 

39.  A  10-kw.    lighting  transformer  takes    122  watts  on  open 
circuit  and  178  watts  on  short  circuit  with  full-load  current  in  the 
windings.     This  transformer  is  connected  to  the  supply  circuit 
continuously.     It  delivers  its  rated  load,   however,  during  but  6 
hours  each  day  and  is  idle  the  remaining  18  hours.     What  is  the 
all-day  efficiency? 

40.  A  lOO-kv.-a.    transformer  supplies  a  lighting    and  power 
load.     The  iron  loss  is  1010  watts  and  the  copper  loss  is  1004  watts 


TRANSFORMERS  9 

with  full-load  current.  This  transformer  is  connected  to  the 
supply  circuit  continuously.  During  7  hours  the  load  is  60  kw- 
at  0.75  power  factor,  and  for  the  remaining  3  hours  of  the  work- 
ing day  the  load  is  80  kw.  at  0.83  power  factor.  The  rest  of  the 
time  no  load  is  supplied.  What  is  the  all-day  efficiency? 

41.  A  oOO-kv.-a.  power  transformer  is  connected  to  the  supply 
circuit  for  12  hours  each  day.  The  iron  loss  is  3330  watts  and 
the  copper  loss  is  4680  watts  with  full-load  current.  During  5 
hours  it  delivers  400  kw.  at  0.80  power  factor  and  during  4  hours, 
250  kw.  at  0.75  power  factor.  The  remaining  hours  it  is  idle. 
What  is  the  all-day  efficiency? 

42. l  The  magnetic  core  of  a  transformer  has  a  mean  length  of 
77.3  in.  and  a  cross-section  of  34.4  sq.  in.  How  many  turns 
should  there  be  in  the  high-tension  winding  if  the  applied  voltage 
is  22,500  volts  at  60  cycles?  The  maximum  flux  density  to  be 
used  is  63,000  lines  per  square  inch,  at  which  the  permeability  of 
the  iron  is  1600.  The  core  loss  per  cubic  inch  of  iron  at  this  flux 
density  and  frequency  is  0.364  watts.  What  are  the  no-load 
high-tension  current  and  power  factor? 

43. l  The  magnetic  core  of  a  25-kv.-a.  transformer  has  a  cross- 
section  of  17.5  sq.  in.  and  an  average  length  of  57.5  in.  The 
high-tension  winding  is  designed  for  an  impressed  voltage  of 
22,000  volts  at  60  cycles  and  for  a  maximum  flux  density  in  the 
magnetic  circuit  of  70,000  lines  per  square  inch.  At  this  flux 
density  the  permeability  is  1860  and  the  iron  loss  is  0.37  watts 
per  cubic  inch.  How  many  turns  should  there  be  in  the  high- 
tension  winding?  What  are  the  no-load  current  and  power 
factor? 

44. l  A  small  experimental  transformer  has  a  magnetic  circuit 
with  a  net  cross-section  of  3.4  sq.  in.  and  a  mean  length  of  22.5 
in.  Assume  that  the  permeability  of  the  iron  is  1800.  There  are 
four  coils,  each  of  which  has  232  turns,  and  a  resistance  of  0.23 
ohm.  (a)  If  all  of  the  coils  are  connected  for  a  2  :1  ratio  of  trans- 
formation what  is  the  magnetizing  component  of  the  no-load 
current  when  220  volts  at  60  cycles  is  impressed  on  the  primary? 
(b)  If  the  eddy  current  and  hysteresis  losses  are  50  watts  what 
is  the  no-load  current?  (c)  What  is  the  efficiency  of  this  trans- 
former when  the  secondary  delivers  20  amperes  at  110  volts  and 
100  per  cent,  power  factor? 

1  Assume  that  at  this  flux  density  75  additional  ampere  turns  are  required 
because  of  the  joints  in  the  magnetic  circuit. 


10     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 


TRANSFORMER  DATA. 


No. 

Kv.-a. 

Type 

Cycles 

Voltage 

Turns 

Magnetic  circuit 

H.T. 

JL.T. 

H.T. 

L.T. 

Area 
sq.  in. 

Av.  length 
in. 

A 
B 
C 
D 

E 
F 

50 
100 
500 
500 
1,000 
3,333 

Core 
Shell 
Core 
Shell 
Shell 
Shell 

60 
60 
60 
25 
60 
60 

22,500 
11,000 
11,000 
13,200 
32,400 
55,000 

2,300 
2,200 
2,300 
425 
11,500 
6,600 

900 

180 

34.4 
66 

88.25 
276 
372 
348 

77.31 
47 
131 
68 
86.8 
92.8 

558 

18 

733 

88 

45. 1  (a)  How  many  turns  should  there  be  in  the  high-  and 
low-tension  windings  in  order  that  the  maximum  flux  density 
shall  be  69,000  lines  per  square  inch  in  transformer  No.  A?  What 
is  the  core  loss?  What  is  the  ratio  of  the  no-load  current  to  the 
full-load  current?  Use  the  magnetic  data  given  in  problem  10. 
(b)  How  many  turns  should  there  be  in  the  high-  and  low-tension 
windings  in  order  that  the  maximum  flux  density  shall  be  69,000 
lines  per  square  inch  in  transformer  No.  C?  What  is  the  core  loss? 
What  is  the  ratio  of  the  no-load  current  to  the  full-load  current? 
Use  the  magnetic  data  given  in  problem  10.  (c)  How  many 
turns  should  there  be  in  the  high-  and  low-tension  windings  in 
order  that  the  maximum  flux  density  shall  be  69,000  lines  per 
square  inch  in  transformer  No.  E?  What  is  the  core  loss?  What 
is  the  ratio  of  the  no-load  current  to  the  full-load  current?  Use 
the  magnetic  data  given  in  problem  10. 

46. 1  (a)  What  are  the  no-load  current  and  core  loss  in  trans- 
former No.  B?  Use  the  magnetic  data  given  in  problem  10.  (b) 
What  are  the  no-load  current  and  core  loss  in  transformer  No.  D? 
Use  the  magnetic  data  given  in  problem  10.  (c)  What  are  the 
no-load  current  and  core  loss  in  transformer  No.  F.?  Use  the 
magnetic  data  given  in  problem  10. 

TRANSFORMER  DATA 


No. 

Kv.-a. 

Type 

Cycles 

Voltage 

Turns 

Magnetic  circuit  |    Open  circuit 

H.T. 

L.T. 

H.T. 

L.T.         Area 

j    sq.  in. 

Av. 

length 
in. 

Watts 

^Volt- 
amps. 

G 
H 
I 

500 
1,000 
5,000 

Shell 
Shell 
Shell 

60 
60 
50 

12,700 
66,000 
52,000 

2,300 
6,600 
33,000 

376 
1,240 
529 

68 
124 
336 

170 
278 
613 

66.4 
95 
140 

3,330 
9,300 
17,500 

25,800 
60,000 
146,000 

47. l  (a)  What  are  the  maximum  flux  density,  the  permeability, 
and  the  iron  loss  in  watts  per  cubic  inch  for  transformer  No.  G? 
(b)  What  are  the  maximum  flux  density,  the  permeability,  and 

1  Assume  that  at  this  flux  density  75  additional  ampere  turns  are  required 
because  of  the  joints  in  the  magnetic  circuit. 


TRANSFORMERS  11 

the  iron  loss  in  watts  per  cubic  inch  for  transformer  No.  H? 
(c)  What  are  the  maximum  flux  density,  the  permeability,  and 
the  iron  loss  in  watts  per  cubic  inch  for  transformer  No.  I? 

48.  A  100-kv.-a.,   11,000  :2200-volt,  60-cycle  transformer  has 
primary  and  secondary  resistances  of  6.0  ohms  and  0.24  ohm 
respectively,  and  primary  and  secondary  leakage  reactances  of 
16  ohms  and  0.64  ohm  respectively.     What  is  the  maximum  per- 
centage change  that  can  occur  in  the  mutual  flux  from  no-load  to 
full-load  current?     At  what  power  factor  would    this  occur? 
Neglect  the  no-load  current. 

49.  A  500-kv.-a.,  11,000  :  2300-volt,  60-cycle  transformer  has 
primary  and  secondary  resistances  of  0.81  ohm  and  0.0358  ohm 
respectively,  and  primary  and  secondary  leakage  reactances  of 
3.7  ohm  and  0.162  ohm  respectively.     With  the  low-tension  wind- 
ing short  circuited  what  voltage  should  be  impressed  on  the  high- 
tension  winding  so  that  the  current  will  have  its  full-load  value? 
Neglect  the  no-load  current. 

50.  A  7.5-kv.-a.,  2080  : 208-volt,  60-cycle  transformer  has  the 
following  constants: 

ri  =  7.53  ohms  #1  =  14. 2  ohms 

r2  =  0.0662  ohm  z2  =  0.128  ohm 

The  core  loss  with  208  volts  impressed  on  the  secondary  is  200 
watts.  Neglect  the  exciting  current,  (a)  What  is  the  regula- 
tion of  this  transformer  for  full-load  current  at  0.8  power  factor? 
(b)  What  is  the  efficiency  of  the  transformer  for  this  load? 

51.  A  50-kv.-a.,  30,000  :  440-volt,  50-cycle  transformer  has  the 
following  constants: 

ri  =  120  ohms  #1  =  428  ohms 

r2  =  0.0258  ohm  x2  =  0.092  ohm 

The  iron  loss  at  the  rated  voltage  is  641  watts.  Neglect  the 
exciting  current,  (a)  What  is  the  regulation  of  this  trans- 
former for  a  load  of  40  kw.,  at  0.75  power  factor?  (b)  What  is 
the  efficiency  of  the  transformer  for  this  load? 

52.  A  500-kv.-a.,  13,200 : 425-volt,  25-cycle   transformer  has 
primary  and  secondary  resistances  of  1.6  ohms  and  0.00166  ohm 
respectively,  and  primary  and  secondary  leakage  reactances  of 
14.5  ohms  and  0.0151  ohm  respectively.     At  no  load  with  425 
volts  impressed  on  the  low-tension  winding  the  power  is  2.5 
kw.  and  the  kilovolt-amperes  are  16.0.     Neglect  any  change  in 


12      PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 


the  exciting  current  with  change  in  the  load.  Calculate  (a)  the 
regulation  and  (b)  the  efficiency  of  this  transformer  when  400 
kw.  is  supplied  to  the  high-tension  winding  at  13,200  volts  and 
0.75  power  factor.  Use  the  complete  transformer  diagram  in  the 
solution  of  this  problem,  (c)  Calculate  the  regulation  and  effi- 
ciency of  this  transformer  for  the  same  load  on  the  assumption 
that  the  no-load  current  is  negligible  and  that  the  core  losses  are 
constant. 

53.     The  following  data  are  given  concerning  a  1000-kv.-a., 
66,000  :  6600-volt,  60-cycle  trafisf ormer : 


No-load 

Short-circuit 

Current  
Voltage  
Power  

9.1 
6600 
9300 

15.  151 
3240 
7490 

Assume  that  the  primary  and  secondary  resistances  are  equal 
when  reduced  to  the  same  side  and  that  the  primary  and  secon- 
dary leakage  reactances  are  similarly  equal.  Assume  that  the 
magnetizing  current  varies  as  the  generated  voltage  and  that  the 
core  loss  varies  as  the  square  of  this  voltage. 

Calculate  (a)  the  regulation  and  (b)  the  efficiency  when  the 
secondary  delivers  full-load  current  at  0.80  power  factor  and  a 
terminal  voltage  of  6680  volts,  (c)  Calculate  the  regulation  and 
efficiency  of  this  transformer  for  the  same  load  on  the  assumption 
that  the  no-load  current  is  negligible  and  that  the  core  losses  are 
constant. 

TRANSFORMER  DATA 


No. 

Kv.-a. 

Type 

Cycles 

Voltage 

Open   circuit 

Short  circuit 

H.T. 

L.T. 

Watts 

Volt- 
amperes 

Volts 

Amperes1!  Watts 

J 

25 

Core 

60 

22,000 

440 

371 

1,700 

1,020 

1.136 

351 

K 

50 

Core 

50 

30,000 

440 

641 

3,000 

1,360 

1.665 

665 

B 

100 

Shell 

60 

11,000 

2,200 

940 

3,750 

310 

9.1 

1,000 

C 

500 

Core 

60 

11,000 

2,300 

2,960 

7,710 

345 

45.5 

3,375 

D 

500 

Shell 

25 

13,200 

425 

2,500 

16,000 

1,100 

37.9 

4,600 

H 

1,000 

Shell 

60 

66,000 

6,600 

9,300 

60,000 

3,240 

15.15 

7,490 

F 

3,333 

Shell 

60 

55,000 

6,600 

13,940 

161,000 

2,350 

60.6 

18,850 

I 

5,000 

Shell 

50 

52,000 

33,000 

17,500 

146,000 

1,800 

151.5 

27,000 

54.  Calculate  (a)  the  regulation2  and  (b)  the  efficiency  of  the 
transformer  No.  J  for  a  load  of  22  kw.  at  0.85  power  factor  and 
rated  voltage. 

1  Full-load  current. 

2  Neglect  the  exciting  current. 


TRANSFORMERS  13 

65.  Calculate  (a)  the  regulation1  and  (b)  the  efficiency  of  the 
transformer  No.  K  for  a  load  of  40  kw  at  0.78  power  factor 
and  rated  voltage. 

56.  Calculate  (a)  the  regulation1  and  (b)  the  efficiency  of  the 
transformer  No.  B  for  a  load  of  100  kw.  at  0.90  power  factor  and 
rated  voltage. 

57.  Calculate  (a)  the  regulation1  and  (b)  the  efficiency  of  the 
transformer  No.  C  for  a  load  of  460  kw.  at  0.91  power  factor 
and  rated  voltage. 

58.  Calculate  (a)  the  regulation1  and  (b)  the  efficiency  of  the 
transformer  No.  D  for  a  load  of  500  kw.  at  0.93  power  factor 
and  rated  voltage. 

59.  Calculate  (a)  the  regulation1  and  (b)  the  efficiency  of  the 
transformer  No.  H  for  a  load  of  1100  kw.  at  unit  power  factor  and 
rated  voltage. 

60.  Calculate  (a)  the  regulation1  and  (b)  the  efficiency  of  the 
transformer  No.  F  for  a  load  of  3500  kw.  at  unit  power  factor 
and  rated  voltage. 

61.  Calculate  (a)  the  regulation1  and  (b)  the  efficiency  of  the 
transformer  No.  I  for  a  load  of  5100  kw.  at  0.92 

power  factor  and  rated  voltage. 

62.  The  resistance  of  the  high-tension  wind- 
ing (a-c)  of  a  2:1  auto  transformer  is  0.072  ohms 
and  that  of  the  low-tension  winding  (b-c)  is  0.035 
ohms.     The  leakage  reactance  of  the  low-tension 
winding  is  0.106  ohm  and  that  of  the  remaining 
portion  of  the  winding  (a-b)  is  0.108  ohm.     If          FlG 
an  electromotive  force  of  12  volts  at  the  rated 

frequency  is  impressed  on  the  high-tension  winding  and  the  low- 
tension  winding  is  short-circuited  what  will  be  the  current  and 
power  taken  from  the  line? 

63.  The   resistance   of    the    high-tension   winding    (a-c) — see 
Fig.  1— of  a  25-kv.-a.,  550  :110-volt,  auto-transformer  is  0.0682 
ohm  and  that  of  the  low-tension  winding  (b-c)  is  0.0042  ohm. 
The  leakage  reactance  of  the  winding  (a-b)  is  0.214  ohm  and  that 
of  the  low-tension  winding  is  0.0132  ohm.     If  the  winding  (a-b) 
is  short-circuited  what  percentage  of  the  rated  voltage  should 
be  applied  to  the  low-tension  winding  (b-c)  in  order  that  full- 
load  current  should  exist  in  the  windings?     What  power  would  be 
supplied? 

1  Neglect  the  exciting  current. 


14     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 


64.  The  resistance  of  the  high-tension  winding  (a-c) — see  Fig. 
1 — of  a  50-kv.-a.,  440  : 110-volt  auto-transformer  is  0.02156  ohm 
and  that  of  low-tension  winding  (6-c)   is  0.00216   ohm.     The 
leakage  reactance  of  the  winding  (a-b)  is  0.068  and  that  of  the 
low-tension  winding  is  0.00756  ohm.     If  the  high-tension  winding 
is  short-circuited,  what  voltage  should  be  impressed  on  the  low- 
tension  winding  in  order  that  there  will  be  full-load  current  in 
the  windings?     What  power  would  be  supplied? 

65.  Calculate  the  regulation  of  the  auto-transformer  described 
in  problem  63  for  a  load  of  22  kw.  at  0.80  power  factor  on  the 
low- tension  side.     If  the  core  loss  at  the  rated  voltage  is  288 
watts  what  is  the  efficiency  of  the  transformer  at  this  load? 

66.  Calculate  the  regulation  of  the  auto-transformer  described 
in  problem  64  for  a  load  of  46  kw.  at  0.87  power  factor  on  the 
low-tension  side.     If  the  core  loss  is  512  watts  at  the  rated 
voltage  what  is  the  efficiency  of  the  transformer  at  this  load? 

AUTO-TRANSFORMER  DATA 


Kv.-a. 

Voltage 

Short  circuit 

Core  loss  at 
rated  veltage 

Watts 

Current* 

Voltsi 

A      1           10 

155:110 

168 

64.5 

7.2 

122 

B 

20 

440-220 

320 

45.5 

20.2 

264 

C 

25 

660:220 

340 

37.9 

29.0 

371 

D 

50 

550:110 

620 

91.0                 25.0 

664 

E 

100 

6,600:440 

940 

15.15             275.0 

980 

F 

500 

11,000:330 

3,850 

45.5               460.0 

2,980 

67.  Calculate  the  regulation  of  the  auto-transformer  No.  A  for 
a  load  of  10  kw.  at  unit  power  factor  on  the  low-tension  side. 
What  is  the  efficiency  of  the  transformer  at  this  load? 

68.  Calculate  the  regulation  of  the  auto -transformer  No.  B 
for  a  load  of  16  kw.  at  0.78  power  factor  on  the  high-tension  side. 
What  is  the  efficiency  of  the  transformer  at  this  load? 

69.  Calculate  the  regulation  of  the  auto-transformer  No.  C 
for  a  load  of  22  kw.  at  0.80  power  factor  on  the  high-tension 
side.     What  is  the  efficiency  of  the  transformer  at  this  load? 

70.  Calculate  the  regulation  of  the  auto-transformer  No.  D. 
for  a  load  of  55  kw.  at  unit  power  factor  on  the  low-tension  side. 
What  is  the  efficiency  of  the  transformer  at  this  load? 

71.  Calculate  the  regulation  of  the  auto-transformer  No.  E 
for  a  load  of  93  kw.  at  0.87  power  factor  on  the  low-tension  side. 
What  is  the  efficiency  of  the  transformer  at  this  load? 

1  The  low-tension  winding  is  short-circuited  and  voltage  is  applied  to  the 
high-tension  winding  so  that  there  is  full-load  current  in  the  windings. 


TRANSFORMERS  15 

72.  Calculate  the  regulation  of  the  auto-transformer  No.  F  for 
a  load  of  510  kw.  at  0.90  power  factor  on  the  low-tension  side. 
What  is  the  efficiency  of  the  transformer  at  this  load? 

73.  If  the  high-tension  winding  of  the  auto-transformer  No.  E  is 
short-circuited  what  voltage  should  be  applied  to  the  low-tension 
winding  so  that  there  will  be  full-load  current  in  the  windings? 

74.  If  the  high-tension  winding  of  the  auto-transformer  No.  F 
is  short-circuited  what  voltage  should  be  applied  to  the  low-ten- 
sion winding  so  that  there  will  be  full-load  current  in  the  windings? 

75.  An   induction  regulator  is  connected  in  a  2300-volt  cir- 
cuit as  shown  in  Fig.  2.     With  no  load  on  the  line  and  with  2300 
volts  impressed  on  the  circuit  as  shown  in  the  ft 

figure  the  voltage  measured  from  a  to  c  is 

2550  volts.     If  the  winding  b-c  is  short-cir-        230QV 

cuited  and  a  reduced  voltage  of  210  volts  is 

impressed  on  the  winding  a-b,  the  line  current      — - —      a 

in  (o-a)  is  130  amperes  and  the  power  supplied  FIG.  2. 

is  2.2  kw.     (a)  When  the  station  supplies  250 

kw.  at  0.78  power  factor  and  a  voltage  of  2300  volts  at  a-b  what  is 

the  line  voltage  at  a-cf     (b)  This  regulator  is  adjusted  to  reduce 

the  voltage  so  that  with  no  load  on  the  line  and  with  2300  volts 

impressed  on  the  circuit  (at  a-b)  the  voltage  at  a-c  is  2050. 

The  short-circuit  data  are  unchanged.     What  is  the  voltage  at  a-c 

when  the  station  supplies  300  kw.  at  a  leading  power  factor  of 

0.92  and  a  line  voltage  of  2300  volts  at  a-b? 

76.  An  induction  regulator  is  connected  in  a  6600-volt  circuit 
as  shown  in  Fig.  2.     With  no  load  on  the  line  and  with  6600  volts 
impressed  on  the  circuit  (at  a-b)  the  voltage  measured  from  a  to  c 
is  7370  volts.     If  the  winding,  b-c,  is  short-circuited,  and  a  reduced 
voltage  of  600  volts  is  impressed  on  the  winding,  a-b,  the  line 
current  (in  o-a)  is  76  amperes  and  power  supplied  is  4.2  kw. 

(a)  When  the  station  supplies  430  kw.  at  0.80  power  factor  and 
a  voltage  of  6600  volts  at  a-b  what  is  the  line  voltage  at  a-cf 

(b)  This  regulator  is  adjusted  to  reduce  the  voltage  so  that 
with  no  load  on  the  line  and  with  6600  volts  impressed  on  the 
circuit  (at  a-b)  the  voltage  at  a-c  is  5850  volts.     The  short- 
circuit  data  are  unchanged.     What  is  the  voltage  at  a-c  when  the 
station  supplies  a  line  current  (in  o-a)  of  81  amperes  at  a  leading 
power  factor  of  0.83  and  a  line  voltage  of  6600  volts  at  a-b? 

77.  A  small  experimental    transformer   has   four  equal  coils 
which  may  be  inter-connected  in  different  ways.     In  the  first  case 


16     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

all  of  the  coils  are  connected  to  form  a  regular  transformer  with  a 
ratio  of  transformation  of  2:1,  and  in  the  second  case  they  are 
connected  to  form  an  auto-transformer  with  the  same  ratio  of 
transformation.  When  connected  as  a  regular  transformer  with 
the  low-tension  winding  short  circuited  the  current,  voltage  and 
power,  measured  on  the  high-tension  side  are  8  amperes,  52  volts, 
and  59  watts  respectively.  On  open  circuit  with  220  volts 
impressed  on  the  high-tension  side  the  core  loss,  is  26  watts. 
Assume  that  the  paths  of  the  leakage  flux  are  the  same  for  each 
connection.  Compare  (a)  the  regulations  and  (b)  the  effi- 
ciencies of  this  transformer  for  the  two  cases  given  above  for  a 
load  of  1.5  kw.  at  0.9  power  factor  and  110  volts  on  the  low-tension 
side. 

78.  Two  transformers,  one  connected   as  a  regular  and  the 
other  as  an  auto-transformer,  have  identical  magnetic  cores  and 
the  same  amount  of  copper  in  their  windings.     Each  of  the  trans- 
formers gives  a  ratio  of  transformation  of  330:220  volts,  and  the 
windings  are  so  designed  that  the  core  losses  in  each  are  122  watts 
when  330  volts  is  impressed  on  their  primaries.     The  primary  and 
secondary  resistances  of  the  regular  transformer  are  0.097  ohm 
and    0.0431    ohm   respectively.     The    resistance    of    the    auto- 
transformer  measured  on  the  low-tension  side  is  also  0.0431  ohm. 
The  leakage  reactance  of  the  primary  winding  of  the  regular 
transformer  is  0.37  ohm.     Assume  that  all  of  the  windings  have 
the  same  mean  length  per  turn,  and  that  the  leakage  reactance  of 
the  coils  vary  as  the  square  of  the  number  of  turns,     (a)  On  the 
basis  of  the  same  total  heating  losses,  what  is  the  ratio  of  the  out- 
puts of  these  two  transformers?     (b)  What  is  the  regulation  and 
efficiency  of  each  transformer  for  a  load  of  10  kw.  at  0.87  power 
factor  and  220  volts? 

79.  Two  transformers,  one  connected  as  a   regular   and  the 
other  as  an  auto-transformer,  have  identical  magnetic  cores  and 
the  same  amount  of  copper  in  their  windings.     Each  of  these 
transformers  gives  a  ratio  of  transformation  of  330:110  volts  and 
the  windings  are  so  designed  that  the  core  losses  of  each  are  371 
watts  when  330  volts  is  impressed  on  their  primaries.     The  pri- 
mary and  secondary  resistances  of  the  regular  transformer  are 
0.0306  ohm  and  0.0034  ohm  respectively.     The  resistances  of  the 
auto-transformer  measured  on  the  high-  and  low-tension  sides  are 
0.017   and  0.0034   ohm  respectively.     The  equivalent  leakage 
reactance  of  the  regular  transformer  referred  to  the  high-tension 


TRANSFORMERS  17 

side  is  0.202  ohm.     Assume  that  the  leakage  reactances  of  the 
windings  vary  as  the  square  of  the  number  of  turns  in  them. 

(a)  If  the  high-tension  winding  of  the  regular  transformer  has 
186  turns  how  many  turns  are  there  in  each  of  the  other  windings 
of  the  regular  and  auto-transformer? 

(b)  The  regular  transformer  is  rated  to  deliver  25  kw.  at  110 
volts.     If  the  losses  at  full  load  are  the  same  for  both  transformers 
what  is  the  kilovolt-ampere  rating  of  the  auto-transformer? 

(c)  What  is  the  regulation  and  efficiency  of  each  transformer 
when  it  is  delivering  its  rated  full-load  current  at  0.83  power  fac- 
tor and  its  rated  voltage? 

80.  Two  transformers,   one  connected  as  a  regular,  and  the 
other  as  an  auto-transformer,  have  identical  magnetic  cores  and 
the  same  amount  of  copper  in  their  windings.     Each  of  these 
transformers  gives  a  ratio  of  transformation  of  550:220  volts  and 
the  windings  are  so  designed  that  the  volts  per  turn  are  the  same 
for  each.     With  25  volts  impressed  on  the  high-tension  winding 
of  the  regular  transformer  and  with  the  low-tension  winding 
short  circuited  the  current  and  power  are  91  amperes  and  665 
watts.     At  the  rated  voltage  the  core  loss  in  the  regular  trans- 
former is  641  watts.     Assume  that  the  resistances  and  leakage 
reactances  of  all  of  the  windings  vary  as  the  square  of  the  number 
of  turns  in  them. 

(a)  The  regular  transformer  is  rated  to  deliver  50  kv.-a.  at 
220  volts.     If   the  losses  at  full  load  are  the  same  for  both 
transformers,  what  is  the  kilovolt-ampere  rating  of  the  auto- 
transformer? 

(b)  What  are  the  regulation  and  efficiency  of  each  transformer 
when  it  is  delivering  its  rated  full-load  power  at  0.9  power  factor? 

81.  Two  transformers,  one  connected  as  a  regular  and  the 
other  as  an  auto-transformer,  have  identical  magnetic  cores  and 
the  same  amount  of  copper  in  their  windings.     Each  of  these 
transformers  gives  a  ratio  of  transformation  of  440:110  volts, 
and  the  windings  are  so  designed  that  the  volts  per  turn  are  the 
same  for  each.     With  14  volts  impressed  on  the  high-tension 
winding  of  the  regular  transformer  and  with  the  low-tension  wind- 
ing short  circuited  the  current  is  227  amperes  and  the  power  is 
1004  watts.     The  core  loss  of  this  transformer  at  rated  impressed 
voltage  is  1010  watts.     Assume  that  the  resistances  and  leakage 
reactances  of  all  of  the  windings  vary  as  the  square  of  the  number 
of  turns  in  them. 


18     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

(a)  The  regular  transformer  is  rated  to  deliver  100  kw.-a.  at 
110  volts.     If  the  losses  at  full   load   are  the  same  for  both 
transformers,  what  is  the  kilovolt-ampere  rating  of  the   auto- 
transformer? 

(b)  What  voltage  should  be  impressed  on  the  high-tension 
winding  of  the  auto-transformer  so  that  there  will  be  full-load 
current  in  the  short-circuited  low-tension  winding? 

(c)  What  are  the  regulation  and  efficiency  of  each  transformer 
for  full-load  current  at  0.8  power  factor? 

82.  A  lO-kv.-a.  and  a  25-kv.-a.  transformer,  each  of  which  has 
a  ratio  of  transformation  of  5:1,  have  their  primaries  connected  in 
parallel    across    an    1100-volt    circuit.     Their    secondaries    are 
also  connected  in  parallel  and  supply  152  amperes  at  0.84  power 
factor  to  an  induction  motor.     Referred  to  the  secondary  sides 
the   equivalent   resistances   are   0.0865   ohm    and   0.0272   ohm 
respectively  and  the  equivalent  reactances  are  0.143  ohm  and 
0.0856  ohm  respectively. 

What  current  does  each  transformer  take  from  the  circuit? 
Compare  these  currents  with  their  rated  full-load  values. 

83.  The  following  short-circuit  data  are  given  on  two  100- 
kv.-a.,  ll,000:460-volt  transformers: 

Type  Amperes1  Voltage  Watts 


Core 
Shell 

9.1 
9.1 

265 
310 

1004 
1000 

These  transformers  are  connected  in  parallel  on  both  the  high- 
and  low-tension  sides  and  supply  a  load  of  186  kw.  at  0.93  power 
factor.  What  percentage  of  its  full-load  current  does  each  trans- 
former supply? 

84.  The  following  short-circuit  data  are  given  on  two  trans- 
formers which  have  a  ratio  of  transformation  of  11,000:2300 
volts: 


Kv.-a. 

Type 

Amperes1 

Volts 

Watts 

100 
500 

Core 
Core 

9.1 
45.5 

265 
345 

1004 
i         3375 

These  transformers  are  connected  in  parallel  on  both  the  high- 
and  low-tension  sides  and  supply  a  total  current  of  293  amperes 
at  0.92  power  factor  on  the  low-tension  side.  What  is  the  current 

1  Full-load  current. 


TRANSFORMERS 


19 


in  each  transformer?     Compare  the  copper  losses  with  their  full- 
load  values. 

85.  -The  following   short-circuit  data  are  given  on  two  500- 
kv.-a.,  ll,000:2300-volt  transformers: 


Type 

Amperes  x 

Voltage 

Watts 

Core 
Shell 

45.5 
45.5 

345 
345 

3375 
4680 

These  transformers  are  connected  in  parallel  on  both  the  high- 
and  low-tension  sides  and  supply  a  total  current  of  452  amperes  at 
0.95  power  factor  on  the  low-tension  side.  What  current  does 
each  transformer  supply? 

86.  The  following  short-circuit  data  are  given  on  three 
66,000 :6600-volt  transformers: 


No. 

Kv.-a. 

Type 

Amperes1 

Volts          |        Watts 

A 
B 
C 

1000 
3333 
5000 

Shell 
Shell 
Shell 

15.15 
50.5 

75.8 

3240 

2820 
3600 

7,490 
18,850 
27,000 

These  transformers  are  connected  in  parallel  on  both  their 
high-  and  low-tension  sides,  and  supply  a  load  of  9200  kw.  at 
a  power  factor  of  0.93  and  their  rated  voltage.  Compare  the 
division  of  the  total  current  between  the  transformers  with  their 
ratings. 

87.  The  following  short-circuit  data  are  given  on  three 
ll,000:460-volt  transformers: 


Kv.-a. 

Type 

Amperes  l 

Volts 

Watts 

100 
500 
500 

Core 
Shell 
Core 

-    9.1 
45.5 
45.5 

265 
917 
345 

1004 
4600 
3375 

These  transformers  are  connected  in  parallel  on  both  the  high- 
and  low-tension  sides,  and  supply  a  load  of  754  kw.  at  0.88  power 
factor  and  their  rated  voltages.  What  current  does  each  trans- 
former supply?  Compare  the  copper  loss  in  each  transformer 
with  the  copper  loss  at  full  load. 

88.  Consider  the  transformers,  Nos.  B  and  C,  described  in 
problem  86.  The  high-tension  windings  receive  power  from  the 
same  66,000-volt  circuit,  while  the  low-tension  windings  deliver 
power  to  independent  circuits.  Each  transformer  delivers  its 

1  Full-load  current. 


20     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 


rated  full-load  current,  the  first  at  unit  power  factor  and  the  sec- 
ond at  0.88  power  factor.  If  the  low-tension  circuits  are  now 
connected  in  parallel  what  is  the  current  in  each  transformer? 
Compare  these  currents  with  their  full-load  values. 

Assume  that  the  currents  taken  by  the  low-tension  circuits  and 
the  power  factors  at  which  they  operate  remain  unchanged. 

89.  Two  50-kw.  transformers  are  connected  in  parallel  on 
both  the  high-  and  low-tension  sides.  Their  constants  are  given 
in  the  following  table: 


Open-circuit  voltage 

High  tension 

Low  tension 

High  tension 

Low  tension 

1*1 

xi                r2 

X2 

22,500 
22,400 

2310 
2320 

61.6 
61.6 

110.            0.661 
110.            0.661 

1.16 
1.16 

They  are  alike  except  for  the  difference  in  their  ratios  of  trans- 
formation. These  transformers  supply  a  combined  load  of  93 
kw.  at  0.89  power  factor  on  the  low-tension  side  at  a  terminal 
voltage  of  2300  volts.  What  is  the  current  in  each  transformer? 

90.  The  following  data  are  given  on  two  transformers  which 
are  operating  with  both  their  high-  and  low-tension  windings 
in  parallel: 


Kv.-a. 

Type 

Rated    open-circuit    voltage 

Short  circuit 

High  tension 

Low  tension 

Amperes1 

Volts 

Watts 

100 
500 

Core 
Shell 

11,000 
13,200 

440 
520 

9.1 
37.9 

265 
1,100 

1,004 
4,600 

The  ratios  of  transformation  are  slightly  different.  These  trans- 
formers take  a  combined  power  of  460  kw.  at  0.90  power  factor 
from  an  11,000-volt  circuit. 

(a)  What  is  the  current  in  each  transformer? 

(b)  If  the  ratios  of  transformation  are  made  equal  by  the 
removal  of  a  few  turns  from  the  high-tension  winding  of  the  500- 
kw.  transformer,  what  is  the  current  taken  by  each  transformer 
for  this  load?     Assume  that  the  short  circuit  data  would  be 
unchanged. 

91.  The  following  data  are  given  on  two  transformers  which 
are  operating  in  parallel  on  both  the  high-  and  low-tension  sides : 


Kv.-a. 

Type 

Rated  open-circuit  voltage 

Short  circuit 

High  tension    |    Low  tension 

Amperes1  |       Volts 

Watts 

500 
500 

Core 
Shell 

13,200 
12,700 

2,340 
2,170 

37.9 
39.4 

413 
332 

3,375 
4,680 

1  Full-load  current  at  rated  voltages. 


TRANSFORMERS  21 

The  ratios  of  transformation  are  slightly  different.  These 
transformers  take  a  combined  power  of  960  kw.  at  0.92  power 
factor  from  a  12,800-volt  circuit. 

(a)  What  is  the  current  in  each  transformer? 

(b)  If  the  ratios  of  transformation  are  made  equal  by  removing 
a  few  turns  from  the  high-tension  winding  of  the  shell-type  trans- 
former what  is  the  current  taken  by  each  transformer  for  this 
load?     Assume  that  the  short-circuit  data  would  be  unchanged. 

92.  Consider    the    transformers    described    in    problem    82. 
They  are  operating  in  parallel  on  both  the  high-  and  low-tension 
sides.     What  are  the  least  values  of  resistance  and  reactance  that 
should  be  added  to  each  transformer  on  the  low-tension  side  in 
order  that  the  currents  supplied  by  the  transformers  shall  be  in 
phase  and  in  proportion  to  their  capacities? 

(b)  For  a  given  load  on  the  transformers  compare  the  total 
heating  in  the  transformers  and  reactors  with  that  in  the  trans- 
formers alone  before  the  addition  of  the  reactors. 

93.  Consider   the   transformers    described   in   problem    83. 
They  are  operating  in  parallel  on  both  the  high-  and  low-tension 
sides. 

(a)  Compare  the  sum  of  their  rated  outputs  with  the  kilovolt- 
ampere  load  they  can  deliver  without  overloading  either  of  them. 

(b)  A  reactor  of  negligible  resistance  is  added  on  the  low- 
tension  side  of  one  transformer  of  such  a  value  that  the  resultant 
impedance  volts  of  the  transformer  and  reactor  on  short  circuit 
with  full-load  current  is  the  same  as  that  of  the  other  transformer 
alone  when  it  also  carries  full-load  current.     What  is  the  imped- 
ance of  this  reactor?     By  what  amount  can  the  combined  output 
of  the  transformers  be  increased  by  the  addition  of  this  reactor 
without  overloading  either  transformer? 

94.  Consider    the   transformers    described    in   problem   84. 
They  are  operating  in  parallel  on  both  the  high-  and  low-tension 
sides. 

(a)  Compare  the  sum  of  their  rated  outputs  with  the  greatest 
kilovolt-ampere  load  they  can  deliver  without  overloading  either 
of  them. 

(b)  A  reactor  of  negligible  resistance  is  added  on  the  low-ten- 
sion side  of  one  transformer  of  such  a  value  that  the  resultant 
impedance  volts  of  the  transformer  and  reactor  on  short  circuit 
with  full-load  current  is  the  same  as  that  of  the  other  transformer 
alone  when  it  also  carries  full-load  current.     What  is  the  imped- 


22     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

ance  of  this  reactor?  By  what  amount  can  the  combined  output 
of  the  transformers  be  increased  by  the  addition  of  this  reactor 
without  overloading  either  transformer? 

95.  The  following  data  are  given  on  two  transformers  which 
are  operating  in  parallel  on  both  their  high-  and  low-tension 
sides : 


Kv.-a. 

Type 

Voltage 

Short  circuit 

High  tension        Low  tension 

Volts 

Amperes*  |      Watts 

1,000 
3,333 

Shell 
Shell 

66,000 
66,000 

6,600 
6,600 

3,360 
2,780 

15.15 
50.5 

7,490 
18,850 

These  transformers  are  delivering  a  combined  load  of  4333 
kv.-a.  at  6600  volts.  A  reactor  of  negligible  resistance  is  added 
in  series  with  one  of  these  transformers  on  the  low-tension  side 
so  that  they  divide  this  load  in  proportion  to  their  ratings.  As- 
sume that  this  does  not  effect  the  low-tension  voltage. 

(a)  What  per  cent,  of  its  rated  capacity  is  the  load  on  each 
transformer  before  the  reactor  is  added? 

(b)  What  is  the  ractance  of  this  reactor?     By  what  amount  is 
the  total  copper  loss  in  the  transformers  reduced? 

96.  Two  similar  5-kw.  lighting  transformers  which  give  a 
ratio  of  transformation  of  1100:110  volts  are  connected  in  series 
on  both  the  high-  and  low-tension  sides.     With  both  low-tension 
windings  short  circuited  and  with  326  volts  impressed  on  the  high- 
tension  windings  connected  in  series  the  current  is  4.55  amperes 
and  the  power  supplied  is  101  watts.     On  the  low-tension  side 
these  transformers  supply  power  to  a  three-wire  system.     The 
resistances  of  the  lamp  loads  on  the  two  sides  of  the  neutral  are 
2.8  ohms  and  2.1  ohms.     The  high-tension  line  voltage  is  2200 
volts.     What  is  the  current  in  the  neutral  conductor?     What  is 
the  voltage  across  each  lamp  load?     If  each  lamp  takes  approxi- 
mately 50  watts  how  many  should  be  turned  off  on  one  side  in 
order  that  all  of  the  lamps  shall  burn  with  the  same  brilliancy? 

97.  The  transformers  described  in  problem  96  are  connected 
in  series  on  both  the  high-  and  low-tension  sides,  and  are  deliver- 
ing power  to  a  three-wire  circuit  on  each  side  of  which  there  are 
two  equal  lamp  loads  taking  32  amperes  at  110  volts. 

What  is  the  high-tension  line  voltage?  If  one  of  these  loads  is 
short  circuited  what  will  be  the  voltage  across  the  other  on  the 
assumption  that  its  resistance  and  the  high-tension  line  voltage 
are  both  unchanged? 

1  Full-load  current. 


TRANSFORMERS  23 

98.  The  transformers  described  in  problem  96  are  connected 
in  parallel  on  the  high-tension  and  in  series  on  the  low-tension 
sides  and  are  delivering  power  to  a  three-wire  lighting  circuit. 
The  resistances  of  the  lamp  loads  on  the  two  sides  of  the  system 
are  2.8  and  2.1  ohms.  What  is  the  current  in  the  neutral  conduc- 
tor? If  the  neutral  conductor  is  disconnected  from  the  trans- 
formers what  is  the  voltage  across  each  lamp  load? 


CHAPTER  II 
SYNCHRONOUS  GENERATORS 

1.  With  normal  excitation  the  resultant  air-gap  flux  in  a  60- 
cycle  alternator  is  106  lines  per  pole.     The  flux  density  is  con- 
stant under  the  pole  face  and  is  zero  between  the  poles. 

If  the  ratio  of  pole  arc  to  pole  pitch  is  0.75  and  the  coil  pitch 
is  1.0,  what  is  the  generated  armature  voltage  per  turn?  What 
is  the  form  factor  of  this  e.m.f.?  Sketch  the  graphs  of  the 
flux  density  and  the  e.m.f. 

2.  In  problem  1  if  the  ratio  of  pole  arc  to  pole  pitch  is  0.75  and 
the  coil  pitch  0.67,  what  is  the  generated  armature  voltage  per 
turn?     What  is  the   form   factor  of    this    e.m.f.?     Sketch  the 
graphs  of  the  flux  density  and  the  e.m.f. 

3.  In  problem  1  if  the  ratio  of  pole  arc  to  pole  pitch  is  0.5  and 
the  coil  pitch  is  1.0,  what  is  the  generated  armature  voltage  per 
turn?     What,  is  the  form   factor  of   this   e.m.f.?     Sketch   the 
graphs  of  the  flux  density  and  the  e.m.f. 

4.  With  normal  excitation  the  resultant  air-gap  in  a  60-cycle 
alternator  is  106  lines  per  pole.     The  flux  density  is  constant  under 
the  pole  face  and  decreases  uniformly  to  zero  at  points  midway 
between  the  poles. 

If  the  ratio  of  pole  arc  to  pole  pitch  is  0.75  and  the  coil  pitch 
is  1.0,  what  is  the  generated  armature  voltage  per  turn?  What 
is  the  form  factor  of  this  e.m.f.?  Sketch  the  graphs  of  the  flux 
density  and  the  e.m.f. 

5.  In  problem  4  if  the  ratio  of  pole  arc  to  pole  pitch  is  0.75 
and  the  coil  pitch  is  0.67,  what  is  the  generated  armature  vol- 
tage per  turn?     What  is  the  form  factor  of  this  e.m.f.?     Sketch 
the  graphs  of  the  flux  density  and  the  e.m.f. 

6.  In  problem  4  if  the  ratio  of  pole  arc  to  pole  pitch  is  0.5  and  the 
coil  pitch  is  1.0,  what  is  the  generated  armature  voltage  per  turn? 
What  is  the  form  factor  of  this  e.m.f.?     Sketch  the  graphs  of  the 
flux  density  and  the  e.m.f. 

7.  With  normal  excitation  the  resultant  air-gap  flux  in  a  60- 
cycle  alternator  is  106  lines  per  pole.     This  flux  is  sinusoidally 
distributed  along  the  air-gap. 

24 


S  YNCHRONO  US  GENERA  TORS  25 

If  the  coil  pitch  is  1.0,  what  is  the  generated  armature  voltage 
per  turn?     What  is  the  form  factor  of  this  e.m.f.? 

8.  In  problem  7  if  the  coil  pitch  is  0.8,  what  is  the  generated 
armature  voltage  per  turn?     What   is  the   form   factor  of  this 
e.m.f.? 

9.  In  problem  7  if  the  coil  pitch  is  0.67,  what  is  the  generated 
armature  voltage  per  turn?     What  is  the  form  factor  of  this 
e.m.f.? 

10.  With  normal  excitation  the  resultant  air-gap  flux  in  a 
60-cycle  alternator  is  106  lines  per  pole.     The  equation  of  the 
curve  which  represents  the  flux  density  in  the  air-gap  is: 

B  =  Bi  sin  x+Ba  sin  3x. 


x  is  the  electrical  angle  measured  from  the  point  midway  between 
the  poles.     Take  B3  =  0.3#i. 

If  the  coil  pitch  is  1.0,  what  is  the  generated  armature  voltage 
per  turn?     What  is  the  form  factor  of  this  e.m.f.? 

11.  In  problem  10  if  the  coil  pitch  is  0.8,  what  is  the  generated 
armature  voltage  per  turn?     What  is  the  form  factor  of  this 
e.m.f.? 

12.  In  problem  10  if  the  coil  pitch  is  0.67  what  is  the  generated 
armature  voltage  per  turn?     What  is  the  form   factor  of  this 
e.m.f.? 

13.  With  normal  excitation  the  resultant  air-gap  flux  in   a 
60-cycle  alternator  is  106  lines  per  pole.     The  equation  of  the 
curve  which  represents  the  flux  density  in  the  air  gap  is: 

B  =  Bl  si 

x  is  the  electrical  angle  measured  from  the  point  midway  between 
the  poles.     Take  B3  =  0.35i. 

If  the  coil  pitch  is  1.0,  what  is  the  generated  armature  voltage 
per  turn?     What  is  the  form  factor  of  this  e.m.f.? 

14.  With  normal  excitation  the  resultant  air-gap  flux  in  a  60- 
cycle  alternator  is  106  lines  per  pole.     The  equation  of  the  curve 
which  represents  the  flux  density  in  the  air-gap  is: 

B  =  Bi  sin  x-\-Bs  sin  (3#  +TT) 

x  is  the  electrical  angle  measured  from  the  point  midway  between 
the  poles.     Take  B3  =  0.3#i. 


26     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

If  the  coil  pitch  is  1.0  what  is  the  generated  armature  voltage 
per  turn?     What  is  the  form  factor  of  this  e.m.f.? 

15.  With  normal"  excitation  the  resultant  air-gap  flux  in  a  60- 
cycle  alternator  is  106  lines  per  pole.     The  equation  of  the  curve 
which  represents  the  flux  density  in  the  air-gap  is: 

B  —  Bi  sin  x-\-Bs,  sin  5x 

x  is  the  electrical  angle  measured  from  the  point  midway  between 
the  poles.     Take  Bb  =  0.2#i. 

If  the  coil  pitch  is  1.0,  what  is  the  generated  armature  voltage 
per  turn?     What  is  the  form  factor  of  this  e.m.f.? 

16.  In  problem  15  if  the  coil  pitch  is  0.8,  what  is  the  generated 
armature  voltage  per  turn?     What  is  the  form  factor  of  this 
e.m.f.? 

17.  In  problem  15  if  the  coil  pitch  is  0.67  what  is  the  generated 
armature  voltage  per  turn?     What  is  the  form  factor  of  this 
e.m.f.? 

18.  With  normal  excitation  the  resultant  air-gap  flux   in,  a 
60-cycle  alternator  is  106  lines  per  pole.     The  equation  of  the 
curve  which  represents  the  flux  density  in  the  air-gap  is: 

B  =  Bi  sin  x—B$  sin  5x 

x  is  the  electrical  angle  measured  from  the  point  midway  between 
the  poles. 

If  the  coil  pitch  is  1.0,  what  is  the  generated  armature  voltage 
per  turn?     What  is  the  form  factor  of  this  e.m.f.? 

19.  With  normal  excitation  the  resultant  air-gap  flux  in  a  60- 
cycle  alternator  is  106  lines  per  pole.     The  flux  density  is  con- 
stant under  the  pole  face  and  is  zero  between  the  poles.     The 
armature  has  8  equally  spaced  slots  per  pole.     The  ratio  of  pole 
arc  to  pole  pitch  is  0.67  and  the  coil  pitch  is  1.0. 

(a)  If  the  armature  winding  consists  of  2  inductors  in  series 
per  pole  placed  in  adjacent  slots,  what  is  the  generated  armature 
voltage  per  pole?     What  is  the  form  factor  of  this  e.m.f.  ?     Sketch 
the  graphs  of  the  flux  density  and  the  e.m.f. 

(b)  Compare  this  e.m.f.   and  form  factor  with  what  they 
would  have  been  had  the  winding  been  concentrated — i.e.,  with 
2  inductors  in  series  per  pole,  placed  in  the  same  slot. 

20.  In  problem  19  if  the  armature  winding  consists  of  4  induc- 
tors in  series  per  pole  placed  in  adjacent  slots,  what  is  the  gene- 


SYNCHRONOUS  GENERATORS  27 

rated  armature  voltage  per  pole?     What  is  the  form  factor  of  this 
e.m.f.?     Sketch  the  graphs  of  the  flux  density  and  the  e.m.f. 

(b)  Compare  this  e.m.f.  and  form  factor  with  what  they  would 
have  been  had  the  winding  been  concentrated — i.e.,  with  4  induc- 
tors in  series  per  pole  placed  in  the  same  slot. 

21.  In  problem  19  if  the  armature  winding  consists  of  4  induc- 
tors in  series  per  pole  placed  in  alternate  slots,  what  is  the  gene- 
rated armature  voltage  per  pole?     What  is  the  form  factor  of 
this  e.m.f.?     Sketch  the  graphs  of  the  flux  density  and  the  e.m.f. 

(b)  Compare  this  e.m.f.  and  form  factor  with  what  they  would 
have  been  had  the  winding  been  concentrated — i.e.,  with  4  induc- 
tors in  series  per  pole  placed  in  the  same  slot. 

22.  With  normal  excitation  the   resultant  air-gap  flux  in  a  60- 
cycle  alternator  is  106  lines  per  pole.     The  flux  density  is  con- 
stant under  the  pole  face  and  decreases  uniformly  to  zero  at  points 
midway  between  the  poles.     The  armature  has  8  equally  spaced 
slots  per  pole.     The  ratio  of  pole  arc  to  pole  pitch  is  0.67  and  the 
coil  pitch  is  1.0. 

If  the  armature  winding  consists  of  2  inductors  in  series  per 
pole  placed  in  adjacent  slots,  what  is  the  generated  armature 
voltage  per  pole?  What  is  the  form  factor  of  this  e.m.f.? 
Sketch  the  graphs  of  the  flux  density  and  the  e.m.f. 

23.  With    normal  excitation  the   resultant  air-gap  flux  in  a 
60-cycle  alternator  is  106  lines  per  pole  and  is  sinusoidally  dis- 
tributed along  the  air-gap.     The  armature  has  12  equally  spaced 
slots  per  pole.     The  coil  pitch  is  1.0. 

(a)  If  the  armature  winding  consists  of  6  inductors  in  series 
per  pole  placed  in  alternate  slots,  what  is  the  generated  armature 
voltage  per  pole?     What  is  the  form  factor  of  this  e.m.f.? 

(b)  Compare  this  e.m.f.  with  what  it  would  have  been  had  the 
6  inductors  per  pole  been  concentrated  in  one  slot. 

24.  In  problem  23  if  the    armature   winding  consists  of  12 
inductors  in  series  per  pole,  one  in  each  slot,  what  is  the  generated 
armature  voltage  per  pole?     What  is  the  form  factor  of  this 
e.m.f.? 

(b)  Compare  this  e.m.f.  with  what  it  would  have  been  had  the 
12  inductors  per  pole  been  concentrated  in  one  slot. 

25.  In  problem  23  if    the  armature    winding  consists  of  8 
inductors  in  series  per  pole  placed  in  adjacent  slots,  what  is  the 
generated  armature  voltage  per  pole?     What  is  the  form  factor 
of  this  e.m.f.? 


28      PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

(b)  Compare  this  e.m.f.  with  what  it  would  have  been  had  the 
8  inductors  per  pole  been  concentrated  in  one  slot. 

26.  With  normal  excitation  the  resultant  air-gap  flux  in  a  60- 
cycle  alternator  is  106  lines  per  pole.  The  equation  of  the  curve 
which  represents  the  flux  density  in  the  air-gap  is 


=    i  sn  x--z  sn    x 

x  is  the  electrical  angle  measured  from  a  point  midway  between 
the  poles.     Take  B3  =  0.3#i. 

(a)  If  the  armature  winding  consists  of  3  equally  spaced 
inductors  per  pole  connected  so  as  to  give  a  winding  which  has  a 
spread  of  one-third  the  pole  pitch  and  a  coil  pitch  of  1.0,  what  is 
the  generated  armature  voltage  per  pole?  What  is  the  form 
factor  of  this  e.m.f.?  (b)  Compare  this  e.m.f.  and  form  factor 
with  what  they  would  have  been  had  the  winding  been  concen- 
trated in  one  slot  per  pole. 

27.  A  single-phase  turbo-alternator   is  rated  to  deliver  1000 
kv.-a.  "at  5200  volts  when  driven  at  1500  rev.  per  min.     The 
armature  has  48  slots,  8  of  which,  or  2  per  pole,  carry  no  inductors, 
so  that  the  spread  of  the  winding  is  0.83.     In  each  of  the  other 
40  slots  there  are  4  inductors  in  series.     What  is  the  no-load 
terminal  voltage  when  the  air-gap  flux  is  40  megalines  per  pole 
and  the  speed  is  1510  rev.  per  min.? 

28.  A  single-phase   alternator  is  rated   to  deliver  250  kv.-a. 
at  2200  volts  when  driven  at  375  rev.  per  min.     The  field  struc- 
ture has  24  poles  and  the  armature,  240  slots.     The  spread  of  the 
armature  winding  is  0.6  and  all  of  the  inductors  are  connected  in 
series.     The  coil  pitch  is  one.     In  any  belt  each  of  the  four  cen- 
tral slots  contains  3  inductors  while  each  of  the  outer  slots  con- 
tains 2  inductors.     What  is  the  no-load  terminal  voltage  when  the 
air-gap  flux  is  3.9  megalines  per  pole  and  the  speed  is  370  rev.  per 
min.? 

29.  A  two-phase  water  turbine  driven  generator  is  rated  to 
deliver  900  kv.-a.  at  5000  volts.     The  field  structure  has  46  poles 
and  the  armature  has  2  slots  per  pole  per  phase.     There  are  11 
inductors  in  series  per  slot.     The  coil  pitch  is  one.     What  is 
the  no-load  terminal  voltage  when  the  air-gap  flux  is  5.4  megalines 
per  pole  and  the  generator  is  driven  at  its  normal  speed  of  120 
rev.  per  min.? 

30.  A  2-phase,  60-cycle,  engine  driven  alternator  is  rated  to 
deliver  3500  kv.-a.   at   11,500  volts.     The   armature   has  576 


S  YNCHRONO  US  GENERA  TORS  29 

slots  with  5  inductors  in  series  per  slot.  The  coil  pitch  is  one. 
The  field  structure  has  96  poles.  What  is  the  generated  armature 
voltage  when  the  resultant  air-gap  flux  is  6.3  megalines  per  pole 
and  the  speed  is  74  rev.  per  min.? 

31.  The  armature  of  a  3-phase,   50-cycle,  12-pole  alternating 
current  generator  has  54  slots  with  2  inductors  per  slot.     The 
inductors  are  arranged  symmetrically  in  the  following  order,  which 
repeats  itself  for  every  pair  of  poles,  i.e.,  for  every  9  slots.     In  the 
first,  third,  fourth,  sixth,  seventh  and  ninth  slots  both  inductors 
are  in  phases  1,  3,  2,  1,  3  and  2  respectively.     In  the  second, 
fifth  and  eighth  slots  the  top  inductors  are  in  phases  3,  1,  and  2 
respectively,  and  the  bottom  inductors  are  in  phases  1,  2  and 
3  respectively.     The  inductors  in  each  phase  are  connected  in 
series  and  the  phases  are  connected  in  Y. 

(a)  If  the  resultant  no-load  air-gap  flux  is  sinusoidally  distrib- 
uted and  has  a  value  of  8X 105  lines  per  pole  what  is  the  terminal 
voltage? 

(b)  What  are  the  reduction  factors  for  the  fifth  and  seventh 
harmonics  in  the  phase  voltage? 

(c)  For  what  harmonics  are  the  reduction  factors  zero? 

32.  The  armature  core  of  a  3-phase  alternating-current  gener- 
ator has  12  slots  per  pole.     Each  slot  contains  2  coil  sides,  so  that 
while  there  are  8  inductors  per  slot  each  coil  has  4  turns.     The 
coil  pitch  is  10  slots. 

What  are  the  total  reduction  factors  for  the  fundamental  and 
the  third,  fifth  and  seventh  harmonics  in  the  phase  voltage? 

33.  A    3-phase,    Y-connected,    25-cycle,    alternating-current 
generator  is  rated  to  deliver  7500  kv.-a  at  a  terminal  potential 
difference  of  12,000  volts.     The  field  structure  has  12  poles  and 
the  armature  has  180  slots.     There  are  4  inductors  in  series  per 
slot.     What  is  the  no-load  terminal  voltage  when  the  air-gap 
flux  is  52  megalines  per  pole  and  the  generator  is  driven  at  its 
rated  speed? 

34.  (a)  In   problem  27,   what    is  the  armature    reaction  in 
ampere  terms  per  pole  when  the  alternator  is  delivering  its  rated 
load  at  unit  power  factor? 

(b)  What  is  the  leakage  reactance  voltage  at  tliis  load?     The 
armature  inductors  are  41  in.  long.     Assume  that  the  leakage 
flux  is  7  lines  per  ampere  per  inch  of  inductor. 

(c)  If  the  armature  winding  had  been  equally  distributed  in 
5  adjacent  slots  instead  of  in  10  what  would  have  been  the  leakage 


30     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

reactance  voltage  at   this  load?     Assume    7    leakage  lines   as 
in(b). 

35.  (a)  In  problem  28  what  is  the  armature  reaction  in  ampere 
turns  per  pair  of  poles  when  the  alternator  is  delivering  200  kw. 
at  0.8  power  factor,  and  its  rated  voltage? 

(b)  What  is  the  leakage  reactance  voltage  at  this  load?     The 
armature  inductors  are  23  in.  long.     Assume  that  the  leakage 
flux  is  8  lines  per  ampere  per  inch  of  inductor. 

(c)  If  the  armature  winding  had  been  equally  distributed  in 
the  four  central  slots,  i.e.,  with  4  inductors  per  slot,  what  would 
have  been  the  leakage  reactance  voltage  at  this  load?     Assume 
8  leakage  lines  as  in  (b) . 

36.  (a)  In  problem  29  what  is  the  armature  reaction  in  ampere 
turns  per  pair  of  poles  when  the  alternator  is  delivering  full-load 
kilo  volt-amperes  ? 

(b)  What  is  the  leakage  reactance  voltage  per  phase  at  this 
load?     Assume  that  the  leakage  flux  is  96  lines  per  ampere  per 
inductor. 

(c)  If  the  armature  winding  had  been  concentrated,  i.e.,  with 
one  slot  per  pole  per  phase,  holding  22  inductors,  what  would 
have  been  the  leakage  reactance  voltage?     Assume  the  same 
leakage  flux  per  inductor  as  in  (b). 

37.  (a)  In  problem  33  what  is  the  armature  reaction  in  ampere 
turns  per  pole  when  the  alternator  is  delivering  7650  kw.  at  0.9 
power  factor  and  its  rated  voltage? 

(b)  What  is  the  leakage  reactance  voltage  per  phase  at  this 
load?  The  armature  inductors  are  48  in.  long.  Assume  that 
the  leakage  flux  is  6.5  lines  per  ampere  per  inch  of  inductor. 

38.  A  three-phase,  Y-connected  alternating-current  generator 
which  is  rated  to  deliver  760  kv.-a.  at  2200  volts  has  an  armature 
with  6  slots  per  pole.     There  are  3  inductors  in  series  per  slot. 

What  are  the  cross-magnetizing  and  demagnetizing  ampere 
turns  per  pole  due  to  armature  reaction  when  the  alternator 
delivers  its  full-load  current  (a)  if  the  power  factor  is  such  that 
the  current  in  any  phase  reaches  its  maximum  value  60  degrees 
after  the  inductors  of  that  phase  pass  the  center  of  the  pole? 
(b)  if  the  power  factor  is  such  that  the  current  in  any  phase 
reaches  its  maximum  value  of  30  degrees  before  the  inductors  of 
that  phase  pass  the  center  of  the  pole? 

39.  A  three-phase,  1500-kv.-a.,  5500-volt  alternating-current 
generator  has  a  field  structure  with  72  poles  and  an  armature 


SYNCHRONOUS  GENERATORS 


31 


with  216  slots.     There  are  12  inductors  in  series  per  slot.     The 
phase  windings  are  connected  in  star. 

(a)  What    are    the    cross-magnetizing    and    demagnetizing 
ampere    turns    per  pole   due  to   armature  reaction  when  the 
alternator  delivers  its  full-load  current  at  0.85  power  factor  (in- 
ductive load)?     Assume  that  for  this  power  factor  the  current 
in  any  phase  is  0.7  of  its  maximum  value  at  the  time  the  in- 
ductors of  that  phase  are  under  the  center  of  the  pole. 

(b)  What  are  the  cross-magnetizing  and  demagnetizing  ampere 
turns  per  pole  due  to  armature  reaction  when  the  alternator 
delivers  its  full-load  current  at  0.85  power  factor  (condensive 
load)?     Assume  that  for  this  power  factor  the  current  in  any 
phase  is  0.2  of  its  maximum  value  at  the  time  the  inductors  of 
that  phase  are  midway  between  the  poles. 

DATA  ON  ALTERNATING-CURRENT  GENERATORS 


A 

B 

Rated  output  (kv.-a.)  

100 

5000 

Phases  .          .... 

2 

3,  Y 

Frequency  

60 

60 

Line  voltage 

480 

6600 

Armature  iron: 
Effective  length  of  core         

5  75  in. 

17  5  in. 

Diameter  of  core  at  air-gap  
Slots  (open)   

26.  4  in. 
96 

148  in. 
360 

Depth  of  slot 

1  488  in. 

2  2  in 

Width  of  slot  

0  37  in. 

0  58  in. 

Field  magnets: 
Poles  

8 

30 

Pole  arc:  Pole  pitch 

0  70 

0  768 

Armature  copper: 
Inductors  per  slot          

8 

4 

Size  of  inductor  

(2  in  parallel) 
No.  8  wire 

(2  in  parallel) 
^X  g-in.  strap 

Thickness  of  insulation  between 
inductor  and  sides  of  slot. 
Mean  length  oer  turn.  . 

0.042  in. 
44.  4  in. 

0.114  in. 
107.  26  in. 

40.  Calculate  the  cross-magnetizing  and  demagnetizing  ampere 
turns  in  alternator  A  for  full-load  current  and  such  a  power 
factor  that  the  current  is  lagging  and  has  a  value  of  0.70  of  its 
maximum  value  in  any  phase  when  that  phase  is  opposite  the 
center  of  the  poles. 

41.  Calculate  the  cross-magnetizing  and  demagnetizing  ampere 
turns  in  alternator  B  for  full-load  current  and  such  a  power  factor 


32     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

that  the  current  is  lagging  and  has  a  value  of  0.80  of  its  maximum 
value  in  any  phase  when  that  phase  is  opposite  the  center  of 
the  poles. 

42.  A  25-cycle  alternating-current  generator  has  an  armature 
core  with  slots  that  are  1 .35  in.  wide  and  2.625  in.  deep.     There  are 
four  inductors  in  series  per  slot,     (a)  Calculate  the  slot  reactance 
per  inch  of  slot  on  the  assumptions  that  the  path  of  the  leakage 
flux  across  the  inductors  is  parallel  to  the  bottom  of  the  slot, 
that  the  permeability  of  the  iron  is  great  and  that  the  inductors 
completely  fill  the  slot. 

(b)  If  the  slot  reactance  is  calculated  on  the  assumption  that 
all  of  the  leakage  flux  per  slot  is  linked  with  all  of  the  inductors 
in  that  slot,  what  value  of  leakage  flux  per  ampere  per  inch  of 
inductor  will  give  the  same  slot  reactance  as  calculated  by  the 
preceding  method? 

43.  A  125-cycle  alternating-current  generator  has  an  armature 
core  with  slots  that  are  1.0  in.  wide  and  3.5  in.  deep.     Each  slot 
holds  12  inductors  in  series  which  completely  fill  it.     (a)  Calculate 
the  leakage  reactance  per  inch  of  slot  on  the  assumption  that  the 
path  of  the  leakage  flux  across  the  inductors  is  parallel  to  the 
bottom  of  the  slot,  and  that  the  permeability  of  the  iron  is  great. 

(b)  If  the  12  inductors  fill  the  slot  to  a  depth  of  but  2.5  in. 
calculate  the  leakage  reactance  per  inch  of  slot  on  the  same 
assumptions. 

44.  A  60-cycle  alternating-current  generator  has  an  armature 
core  with  slots  that  are  1.1  in.  wide  and  2.75  in.  deep  and  15  in. 
long.     There  are  6  inductors  in  series  per  slot  which  fill  it  to  a 
depth  of  2.0  in.     (a)  Calculate  the  slot  reactance  on  the  assump- 
tion that  the  path  of  the  leakage  flux  between  the  sides  of  the 
slot  is  parallel  to  the  bottom  of  the  slot,  and  that  the  permea- 
bility of  the  iron  is  great. 

(b)  If  the  slot  reactance  is  calculated  on  the  assumption  that 
all  of  the  leakage  flux  per  slot  is  linked  with  all  of  the  inductors 
in  that  slot,  what  value  of  leakage  flux  per  ampere  per  inch  of 
inductor  will  give  the  same  slot  reactance  as  calculated  by  the 
preceding  method? 

45.  A    3-phase,   25-cycle   altenator  is  rated  to   deliver  850 
kv.-a.  at  5000  volts.     The  slots  in  the  armature  core  are  1.31 
in.  wide  and  3.625  in.  deep.     The  gross  length  of  the  armature 
core  is  14.5  in.  and  the  effective  length,  deducting  for  the  ven- 
tilating ducts  and  the  insulation  on  the  laminations,  is  10.75 


S  YNCHRONO  US  GENERA  TORS  33 

in.  There  are  14  inductors  in  series  per  slot  which  occupy  a 
space  0.75  in.  by  2.75  in.  The  thickness  of  the  insulation 
between  the  inductors  and  the  armature  core  is  the  same  at  the 
sides  and  at  the  bottom  of  the  slot.  The  mean  length  of  one 
turn  is  93  in.  Each  coil  consists  of  14  turns. 

(a)  Calculate  the  leakage  reactance  per  coil  on  the  following 
assumptions.     Where  the  inductors  are  embedded  in  iron  the 
path  of  the  leakage  flux  between  the  sides  of  the  slot  is  parallel  to 
the  bottom  of  the  slot.     For  the  portions  of  the  coil  that  are 
not  embedded  in  the  armature  core  the  leakage  flux  links  with 
all  of  the  turns  in  that  coil  and  has  a  value  of  0.8  line  per  ampere 
per  inch  of  turn. 

(b)  If  the  leakage  reactance  is  calculated  on  the  assumption 
that  all  of  the  leakage  flux  per  coil  links  with  all  of  the  turns  in 
the  coil,  and  that  the  length  of  the  coil  is  but  twice  the  length  of 
the  inductor,  i.e.,  twice  the  gross  length  of  the  armature  core, 
what  value  of  leakage  flux  per  ampere  per  inch  of  inductor  will 
give  the  same  value  of  leakage  reactance  as  calculated  by  the 
preceding  method? 

46.  (a)  Calculate  the  slot  and  coil  end  leakage  reactance  for 
alternator  A  (see  problem  40).     Use  the  most  exact  method  at 
your  command  for  the  data  given. 

(b)  Calculate  the  slot  and  coil  end  leakage  reactance  for 
alternator  B  (see  problem  41).  Use  the  most  exact  method  at 
your  command  for  the  data  given. 

47.  A  special  60-cycle  generator  has  a  field  structure  with  6 
poles  and  an  armature  with  72  slots.     There  are  9  inductors  in 
series  per  slot.     The  coils  in  adjacent  slots  are  connected  in 
series  by  pairs  so  that  each  of  the  6  independent  armature  wind- 
ings thus  formed  has  a  spread  of  one-sixth  and  a  pitch  of  1.0. 
When  the  6  windings  are  arranged  to  form  a  3-phase,  Y-con- 
nected    armature  winding,  the  synchronous    reactance  voltage 
is  46  volts  for  an  armature  current  of  30  amperes.     Assume 
that  the  armature  leakage  flux  is  72  lines  per  inductor  per  am- 
pere, and  that  the    approximate    formula   for    armature  reac- 

77 
tion,  0.707-p-'  holds  exactly  in  every  case. 

(a)  What  is  the  synchronous  reactance  voltage  with  an  arma- 
ture current  of  30  amperes  when  the  6  armature  windings  are 
connected  in  series  to  form  a  single-phase,  open-circuit  armature 
winding? 


34     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

(b)  The  synchronous  reactance  voltage  is  24  per  cent,  of  the 
open-circuit  phase  voltage  when  the  armature  winding  is  con- 
nected in  Y.  With  the  same  field  current  what  per  cent,  of  the 
open-circuit  armature  voltage  will  the  synchronous  reactance 
voltage  be  when  the  windings  are  connected  to  form  a  single- 
phase,  open-circuit  armature  winding? 

48.  In  problem  47  what  is  the  synchronous  reactance  voltage 
with  an  armature  current  of  60  amperes — i.e.,  30  amperes  per 
inductor — when  the  6  windings  are  connected  to  form  a  single- 
phase,  2-circuit  armature  winding? 

(b)  With  the  same  field  current  as  in  47  (b)  what  per  cent,  of 
the  open-circuit  armature  voltage  will  the  synchronous  react- 
ance voltage  be  when  the  windings  are  connected  to  form  a 
single-phase,  2-circuit  armature  winding? 

49.  In  problem  47  what  is  the  synchronous  reactance  voltage 
with  an  armature  current  of  30  amperes  when  the  6  windings 
are  connected  to  form  a  2-phase  armature  winding? 

(b)  With  the  same  field  current  as  in  47 (b)  what  per  cent,  of 
the  open-circuit  armature  voltage  will  the  synchronous  reactance 
voltage  be  when  the  windings  are  connected  to  form  a  2-phase 
armature  winding? 

60.  Two  three-phase,  60-cycle  alternating-current  generators 
have  the  same  current  and  voltage  rating.  The  first  has  12 
slots  per  pole  and  the  second  has  9  slots  per  pole.  Each  alter- 
nator has  the  same  number  of  turns  per  phase.  The  first  has 
4  poles  and  the  second  6  but  the  dimensions  of  the  magnetic 
circuit  are  such  that  the  same  number  of  ampere  turns  on  the 
field  produces  the  same  flux  per  pole  in  each.  The  shape  of  the 
armature  slots  is  such  that  the  leakage  flux  per  ampere  per 
inductor  is  also  the  same  for  each. 

The  armature  leakage  reactance  and  the  synchronous  react- 
ance of  the  first  alternator  are  respectively  0.54  ohms  and  2.13 
ohms.  What  are  the  corresponding  constants  of  the  second 
alternator? 

51.  A  3-phase,  delta-connected  alternating-current  generator 
is  rated  to  deliver  15  kv.-a.  at  230  volts  when  running  at  a  speed 
of  1200  rev.  per  min.  There  are  6  field  poles  each  of  which  is 
wound  with  398  turns.  The  armature  has  72  slots  with  8-induc- 
tors  in  series  per  slot.  Each  inductor  is  5  in.  long.  The  hot 
resistance  of  the  armature  measured  between  any  two  terminals 
is  0.126  ohm.  The  effective  resistance  is  1.6  times  the  ohmic 


SYNCHRONOUS  GENERATORS 


35 


resistance.  In  calculating  the  leakage  reactance  assume  8 
leakage  lines  per  ampere  per  inch  of  inductor.  The  open-  and 
short-circuit  characteristics  are  given  by  the  following  data: 


Field  current 

Open  circuit 

Short  circuit 

Terminal  voltage 

Armature  current 

2.0 

104 

22 

4.0 

199 

44 

5.0 

240 

55 

6.0 

275 

7.5 

'     315 

(a)  Calculate  the  regulation  of  this  generator  by  the  general 
method  for  an  inductive  load  of  15  kw.  at  0.8  power  factor. 
What  is  the  field  current  calculated  by  this  method? 

(b)  Calculate  the  regulation  of  this  generator  for  the  specified 
load  by  the  synchronous  impedance  method.     What  is  the  field 
current  calculated  by  this  method? 

•  (c)  Calculate  the  regulation  of  this  generator  for  the  specified 
load  by  the  magnetomotive  force  method.  What  is  the  field 
current  calculated  by  this  method? 

52.  A  760-kv.-a,  2200-volt  alternating-current  generator 
delivers  energy  directly  to  a  3-phase,  50-cycle  system.  The 
neutral  pf  the  generator  is  grounded.  The  field  structure 
consists  of  64  poles  each  of  which  is  wound  with  50  turns.  The 
armature  core  has  384  slots  with  3  inductors  in  series  per  slot. 
The  length  of  the  armature  core  parallel  to  the  shaft  is  10  in. 
The  effective  resistance  of  the  armature  winding  is  0.172  ohm  per 
phase.  In  calculating  the  leakage  reactance  assume  6.5  leakage 
lines  per  ampere  per  inch  of  slot  per  inductor.  The  open-  and 
short-circuit  characteristics  are  given  by  the  following  data: 


Open  circuit 

Short  circuit 

Terminal  voltage 

Armature  current 

50 

80 
100 

1060 
1650 
1950 

280 
450 

120 

2160 

150 

2420 

200 

2650 

(a)  Calculate  the  regulation  of  this  generator  for  25  per  cent, 
overload  current  at  unit  power  factor  by  the  general  method. 
What  is  the  field  current  calculated  by  this  method? 


36      PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

(b)  Calculate  the  regulation  of  this  generator  for  the  specified 
load  by  the  synchronous  impedance  method.     What  is  the  field 
current  calculated  by  this  method? 

(c)  Calculate  the  regulation  of  this  generator  for  the  specified 
load  by  the  magnetomotive  force  method.     What  is  the  field 
current  calculated  by  this  method? 

53.  A  3-phase,  Y-connected,  1500-kv.-a.,  5500-volt  alternating- 
current  generator  is  driven  by  a  2000-h.p.  reciprocating  engine 
that  runs  at  83  rev.  per  min.  at  full  load.  The  field  structure 
has  72  poles  each  of  which  is  wound  with  35  turns.  The  armature 
core  has  1  slot  per  pole  per  phase  with  12  inductors  in  series  per 
slot.  Each  inductor  is  9  in.  long.  The  armature  winding  has 
an  effective  resistance  of  0.36  ohm  per  phase.  In  calculating 
the  leakage  reactance  assume  7.5  equivalent  leakage  lines  per 
ampere  per  inch  of  inductor.  The  data  for  the  open-  and  short- 
circuit  characteristics  are: 

Open  circuit  Short  circuit 


rituu  current 
Terminal  voltage           Armature  current 

150 
200 
250 
300 
350 

5100 
5900 
6500 
6800 
7100 

300 
400 

(a)  Calculate  the  regulation  of  this  generator  by  the  general 
method  for  20  per  cent,  overload  current  at  0.85  power  factor 
(induction  load).     What  is  the  field  current  calculated  by  this 
method? 

(b)  Calculate  the  regulation  of  this  generator  for  the  specified 
load  by  the  synchronous  impedance  method.     What  is  the  field 
current  calculated  by  this  method? 

(c)  Calculate  the  regulation  of  this  generator  for  the  specified 
load  by  the  magnetomotive  force  method.     What  is  the  field 
current  calculated  by  this  method? 

54.  A  3750-kv.-a.  alternating-current  generator  delivers  energy 
at  2200  volts  to  a  2-phase,  30-cycle  system.  The  field  structure 
has  20  poles,  each  of  which  is  wound  with  60  turns.  The  arma- 
ture has  360  slots  with  one  inductor  per  slot.  Each  inductor  is 
20.5  in.  long.  The  armature  resistance  by  direct-current  measure- 
ment is  0.0196  ohm  per  phase  at  25°  C.  Assume  that  the  ratio 
of  effective  resistance  to  ohmic  resistance  is  1.3  at  25°  C.  In 


SYNCHRONOUS  GENERATORS 


37 


calculating  the  leakage  reactance  assume  that  the  leakage  flux 
is  6.5  lines  per  ampere  per  inch  of  inductor.  The  data  for  the 
open-  and  short-circuit  characteristics  are: 


Field  ampere  turns 

Open  circuit 

Short  circuit 

per  pole 

Terminal  voltage 

Armature  current 

4,000 

8,000 
12,000 
16,000 

680 
1360 
2000 
2480 

606 
1210 

18.000 

2660 

(a)  Calculate  the  regulation  of  this  generator  for  a  condensive 
load  of  3500  kw.  at  0.92  power  factor  by  the  general  method. 
What  is  the  field  current  calculated  by  this  method?     Assume 
that  the  temperature  of  the  armature  windings  is  70°  C. 

(b)  Calculate  the  regulation  of  this  generator  for  the  specified 
load  by  the  synchronous  impedance  method.     What  is  the  field 
current  calculated  by  this  method? 

(c)  Calculate  the  regulation  of  this  generator  for  the  specified 
load  by  the  magneto-motive  force  method.     What  is  the  field 
current  calculated  by  this  method? 

55.  A  3-phase  water-wheel  generator  whose  armature  winding 
is  Y-connected  is  rated  to  deliver  5000  kv.-a.  at  6600  volts. 
Normal  speed  is  240  rev.  per  min.  The  field  structure  has  30 
poles  with  67.5  turns  per  pole.  The  armature  core  has  360 
slots  with  2  inductors  in  series  per  slot.  The  length  of  the 
armature  core  is  21.5  in.  The  measured  resistance  of  the  arma- 
ture between  any  two  terminals  is  0.0836  ohm  at  25°  C.  The 
ratio  of  effective  resistance  to  ohmic  resistance  is  1.65  at  25°  C. 
In  calculating  the  leakage  reactance  assume  that  the  leakage  flux 
is  6.5  lines  per  ampere  per  inch  of  slot  per  inductor. 

The  open-  and  short-circuit  characteristics  are  given  by  the 
following  data: 


. 

Open  circuit 

Short  circuit 

Terminal  voltage 

Armature  current 

100 
150 
200 

4800 
6500 
7400 

680 
1020 

250 

7900 

(a)  Calculate  the  regulation  of  this  generator  by  the  general 
method  for  full-load  kilovolt-amperes  at  0.8  power  factor  (indue- 


38     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

tive  load).  What  is  the  field  current  calculated  by  this  method? 
Assume  that  the  temperature  of  the  armature  windings  is 
70°  C. 

(b)  Calculate  the  regulation  of  this  generator  for  the  specified 
load  by  the  synchronous  impedance  method.     What  is  the  field 
current  calculated  by  this  method? 

(c)  Calculate  the  regulation  of  this  generator  for  the  specified 
load  by  the  magnetomotive-force  method.     What  is  the  field 
current  calculated  by  this  method? 

56.  A  2-phase,  60-cycle  alternating-current  generator  is  rated 
to  deliver  100  kv.-a.  at  480  volts.  The  armature  has  an  effective 
resistance  of  0.138  ohm  and  a  leakage  reactance  of  0.159  ohm  per 
phase.  When  the  power  factor  of  the  load  is  0.8  the  armature 
demagnetizing  ampere  turns  are  11.8  and  the  cross-magnetizing 
ampere  turns  are  15.2  per  pole  per  ampere.  The  field  poles 
are  each  wound  with  265  turns.  The  data  for  the  open  circuit 
characteristic  are: 


Field  current 

Open-circuit  voltage 

10 
15 
20 
25 

400 
500 
560 

!                                 598 

What  is  the  regulation  of  this  generator  when  delivering  full- 
load  current  at  0.8  power  factor  (inductive  load)  ? 

(a)  Assume   that   the   cross-magnetizing  and   demagnetizing 
ampere  turns  act  on  magnetic  circuits  of  the  same  reluctance  as 
that  of  the  resultant  field.     This  is  a  modification  of  the  general 
method. 

(b)  Assume  that  the  cross-magnetizing  ampere  turns  act  on  a 
magnetic  circuit  whose  reluctance  is  determined  by  the  lower 
part  of  the  saturation  curve,  and  that  the  demagnetizing  ampere 
turns  act  on  the  same  magnetic  circuit  as  do  the  impressed  field 
ampere  turns.     This  is  the  Blondel  method. 

57.  A  3-phase,  25-cycle  alternating-current  generator  is  rated 
to  deliver  850  kv.-a.  at  5000  volts.  The  armature  windings  are 
connected  in  Y,  and  have  an  effective  resistance  of  0.398  ohm  and 
a  leakage  reactance  of  1.46  ohms  per  phase.  When  the  power 
factor  of  the  load  is  unity  the  armature  demagnetizing  ampere 
turns  are  5.8  and  the  cross-magnetizing  ampere  turns  are  31  per 


SYNCHRONOUS  GENERATORS 


39 


pole  per  ampere.     The  field  poles  are  each  wound  with 
turns.     The  data  for  the  open-circuit  characteristic  are: 


79.5 


Field  ampere  turns  per  pole  Open-circuit  terminal  voltage 


6,000 
10,000 
12,000 
14,000 


4330 
5460 
5800 
6060 


What  is  the  regulation  of  this  generator  when  delivering  30 
per  cent,  overload  current  at  unit  power  factor? 

(a)  Assume   that  the   cross-magnetizing   and   demagnetizing 
ampere  turns  act  on  magnetic  circuits  of  the  same  reluctance  as 
that  of  the  resultant  field.     This  is  a  modification  of  the  general 
method. 

(b)  Assume  that  the  cross-magnetizing  ampere  turns  act  on  a 
magnetic  circuit  whose  reluctance  is  determined  by  the  lower 
part  of  the  saturation  curve,  and  that  the  demagnetizing  ampere 
turns  act  on  the  same  magnetic  circuit  as  do  the  impressed  field 
ampere  turns.     This  is  the  Blondel  method. 

58.  A  3-phase,  Y-connected  alternating-current  generator  is 
rated  to  deliver  1000  kv.-a.  at  13,800  volts.  The  armature 
has  an  effective  resistance  of  2.18  ohms  per  phase.  The  data  for 
the  open-circuit  characteristic  and  the  full-load  current  satura- 
tion curve  at  zero  power  factor  are: 


Field  current 

Open-circuit  terminal 
voltage 

Saturation  curve 
42  amp.,  at  zero  P.F. 

41  2 

0 

50 

8,800 

110 
140 
180 

15,600 
17,250 
18,900 

10,750 
13,250 
15,600 

What  is  the  regulation  for  an  inductive  load  of  50  amperes 
at  0.85  power  factor? 

59.  A  3-phase,  water-wheel  generator  is  rated  to  deliver  5000 
kv.-a.  at  6600  volts.  The  armature  winding  is  Y-connected 
and  has  an  effective  resistance  of  0.081  ohm  per  phase.  The  data 
for  the  open-circuit  characteristic  and  the  full-load  current  satura- 
tion curve  at  zero  power  factor  are: 

4 


40     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 


Field  current 


Open-circuit  terminal 
voltage 


Saturation  curve 
438  amp.  at  zero  P.F. 


64 

3050 

0 

100 

4800 

1800 

150 

6500 

4200 

200 

7400 

5700 

275 

8150 

6900 

What  is  the  regulation  for  an  inductive  load  of  500  amperes 
at  0.8  power  factor? 

60.  A  3-phase,  Y-connected  alternating-current  generator  is 
rated  to  deliver  1640  kv.-a.  at  13,500  volts.     The  armature  has 
an  effective  resistance  of  1.52  ohms  per  phase  and  a  synchronous 
reactance  of  37.4  ohms  per  phase. 

(a)  What  is  the  regulation  of  this  alternator  on  an  inductive 
load  taking  1500  kw.  at  0.85  power  factor? 

(b)  What  is  the  regulation  on  a  condensive  load  taking  1500 
kw.  at  0.85  power  factor? 

61.  A  2-phase  alternating-current  generator  is  rated  to  deliver 
3500  kw.  at  10,000  volts.     The  armature  has  an  effective  resist- 
ance of  0.64  ohm  per  phase  and  a  synchronous  reactance  of  23.7 
ohms  per  phase. 

(a)  What  is  the  regulation  of  this  alternator  on  a  non-induc- 
tive load  taking  the  rated  kv.— a.? 

(b)  What  is  the  regulation  on  an  inductive  load  taking  the 
rated  kv.-a.  at  zero  power  factor? 

(c)  What  is  the  regulation  on  a  condensive  load  taking  the 
rated  kv.-a.  at  zero  power  factor? 

62.  A  760-kv.-a,  2200-volt,  3-phase  alternating-current  gen- 
erator has  an  effective  armature  resistance  of  0.17  ohm  per  phase. 
The  armature  winding  is  connected  in  Y.     With  the  armature 
short-circuited  the  armature  current  is  338  amperes  when  the 
field  current    is  60  amperes.     The  open-circuit    characteristic 
data  are: 


Field  current 

Open-circuit  terminal  voltage 

50 
80 
100 
120 
150 
200 

1060 
1650 
1950 
2160 
2420 
2650 

S  YNCHRONO  US  GENERA  TORS  4 1 

(a)  What  is  the  regulation  of  this  generator,  calculated  by  the 
magnetomotive-force  method,  for  an  inductive  load  which  re- 
quires full-load  current  at  0.5  power  factor? 

(b)  What  is  the  regulation  for  a  condensive  load  which  re- 
quires full-load  current  at  0.5  power  factor? 

63.  A    3-phase,    A-connected    alternating-current    generator 
has  a  full-load  capacity  of  15  kv.-a.  at  230  volts.     The  effective 
armature  resistance  and  synchronous  reactance  are  respectively 
0.302  and  4.36  ohms  per  phase.     Three  reactors  each  of  which 
has  an  effective  resistance  of  2.5  ohms  and  a  reactance  of  10 
ohms  are  connected  in  A  across  the  terminals  of  the  generator. 
If  the  terminal  voltage  is  adjusted  to  its  rated  value  to  what  will 
it  rise  when  the  coils  are  removed? 

(b)  If  the  coils  are  connected  in  Y  and  the  terminal  voltage 
adjusted  as  before  to  what  will  it  rise  when  the  coils  are  removed? 

(c)  What  is  the  power  output  of  the  generator  in  each  case? 

64.  In  problem  63  if  the  open-circuit  voltage  of  the  generator 
is  adjusted  to  300  volts  to  what  will  it  fall  when  the  coils  are 
connected  in  A  across  the  terminals?     What  is  the  power  out- 
put of  the  generator? 

65.  A    3-phase,     1500-kv.-a.,    5500-volt    alternating-current 
generator  delivers  full-load  current  to  an  inductive  load  at  0.85 
power  factor.     The  effective  resistance  and  synchronous  react- 
ance of  this  alternator  are  respectively  0.36  and  8.5  ohms  per 
phase.     The  armature  winding  is  Y-connected.     With  the  field 
excitation  unchanged  what  will  be  the  terminal  voltage  if  the 
alternator  delivers  its  rated  current  to  a  condensive  load  at  0.85 
power  factor? 

66.  A  special  6-pole,   60-cycle  alternating-current  generator 
has-    six    similar    and    independent    armature    windings.     The 
windings  are  equally  spaced  so  that  their  voltages  differ  by  30 
degrees. 

On  the  basis  of  equal  armature  and  field  heating  losses  and  of 
equal  frequencies  compare  the  rated  outputs — kilovolt-amperes 
and  terminal  voltage — of  this  alternator  when  the  windings  are 
connected  (1)  for  an  open-coil  single-phase  winding  and  (2)  for  a 
2-phase  winding. 

67.  (a)  In  problem  66  compare  the  rated  outputs  on  the  same 
basis  when  the  windings  are  connected  (1)  for  an  open-coil  single- 
phase  winding  and  (2)  for  a  3-phase  mesh  winding. 

(b)  Compare  the  rated  outputs  on  the  same  basis  when  the 


42     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

windings  are  connected  (1)  for  an  open-coil  single-phase  winding 
and  (2)  for  a  3-phase  star  winding. 

68.  In  problem  66  compare  the  rated  outputs  on  the  same 
basis  when  the  windings  are  connected  (1)  for  a  2-phase  winding 
and  (2)  for  a  3-phase  star  winding. 

69.  Concerning  the  alternator  described  in  problem  51  the 
following  additional  data  are  given.     The  field  current  is  sup- 
plied at  110  volts.     The  friction  and  windage  loss  is  310  watts  at 
normal  speed,  and  the  core  loss  due  to  rotation  is  480  watts  for 
an  armature  generated  voltage  of  240.     Assume  that  the  core 
loss  is  constant.     What  is  the  efficiency  of  this  alternator  at  the 
load  described  in  problem  51?     (1)  Calculate  the  field  current 
by  the  synchronous  impedance  method,   (2)  by  the  magneto- 
motive-force method. 

70.  Concerning  the  alternator  described  in  problem  52  the 
following  additional  data  are  given.     The  resistance  of  the  field 
circuit  is  0.516  ohm.     The  friction  and  windage  loss  is  6.2  kw. 
at  normal  speed.     The  core  loss  due  to  rotation  is  11.0  kw.  at 
2200  volts  and  may  be  assumed  to  vary  as  the  square  of  the 
generated  armature  voltage. 

(a)  What  is  the  efficiency  of  this  generator  at  the  load  de- 
scribed in  problem  52? 

(b)  What  is  the  efficiency  of  this  generator  when  delivering 
the  same  current  at  0.80  power  factor? 

Calculate  the  field  current  (1)  by  the  general  method  and  (2) 
by  the  magnetomotive-force  method. 

71.  Concerning  the  alternator  described  in  problem  53  the 
following   additional   data   are   given.     The   resistance   of   the 
field  circuit  is  0.376  ohm.     The  friction  and  windage  loss  is  8.4 
kw.     The  core  loss  due  to  rotation  is  20.2  kw.  at  5500  volts,  and 
should  be  assumed  to  vary  as  the  square  of  the  generated  arma- 
ture voltage. 

(a)  What  is  the  efficiency  of  the  generator  at  the  load  de- 
scribed in  problem  53? 

(b)  What  is  the  efficiency  of  this  generator  when  delivering 
the  same  power  at  unit  power  factor? 

Calculate  the  field  current  (1)  by  the  synchronous  impedance 
method,  and  (2)  by  the  magnetomotive-force  method. 

72.  The  full-load  capacity  of  a  3-phase,  25-cycle  alternating- 
current  generator  is  850  kv.-a.  at  5000  volts.     The  armature 
windings  are  connected  in  Y  and  have  an  effective  resistance  of 


S  YNCHRONO  US  GENERA  TORS  43 

0.398  ohm  per  phase.  The  synchronous  reactance  is  15.4  ohms 
per  phase.  With  the  armature  short-circuited  the  armature 
current  is  108  amperes  when  the  field  current  is  50  amperes. 
The  resistance  of  the  field  circuit  is  0.82  ohm.  The  friction  and 
windage  is  7.6  kw.  The  core  loss  due  to  rotation  is  20.2  kw. 
at  5200  volts  and  may  be  assumed  to  be  constant.  The  open- 
circuit  characteristic  is  given  by  the  following  data: 

Field  current  Open-circuit  terminal  voltage 


125 
150 
175 


5460 
5800 
6060 


What  is  the  efficiency  of  this  generator  when  delivering  10  per 
cent,  overload  current  at  0.80  power  factor  (inductive)? 

Calculate  the  field  current  (1)  by  the  synchronous  impedance 
method,  and  (2)  by  the  magentomotive  force  method. 

73.  Concerning  the  alternator  described  in  problem  54  the 
following  additional  data  are  given.  The  resistance  of  the  field 
circuit  is  0.37  ohm.  The  friction  and  windage  is  19  kw.  The 
core  loss  is  given  by: 


Terminal  voltage  on  open  circuit  Core  loss 


1360 
2000 
2480 


18.0  kw. 
36.5  kw. 
60.0  kw. 


What  is  the  efficiency  of  this  generator  for  the  load  described 
in  problem  54?  Calculate  the  field  current  (1)  by  the  general 
method,  (2)  by  the  synchronous  impedance  method,  and  (3)  by 
the  magnetomotive-force  method. 

74.  Concerning  the  alternator  described  in  problem  58  the 
following  additional  data  are  given.  The  resistance  of  the  field 
circuit  is  0.541  ohm.  The  friction  and  windage  loss  is  9.2  kw. 
The  core  loss  is  given  by: 

Terminal  voltage  on  open  circuit  |  Core  loss 


8,800 
13,000 
15,600 
17,250 


7.5  kw. 
16.6  kw. 

25.4  kw. 

33.5  kw. 


What  is  the  efficiency  of  this  generator  for  the  load  described 
in  problem  58? 


44     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

75.  Concerning  the  alternator  described  in  problem  59  the 
following  additional  data  are  given.  The  resistance  of  the  field 
circuit  is  0.549  ohm  at  25°  C.  The  temperature  of  the  field 
under  load  conditions  is  68°  C.  The  friction  and  windage  loss 
is  38  kw.  The  core  loss  is  given  by: 


Terminal  voltage  on  open  circuit 

Core  loss 

4800 
6000 
6600 
7500 

45  kw. 
73  kw. 
90  kw. 
123  kw. 

What  is  the  efficiency  of  this  generator  for  the  load  described  in 
problem  59? 

76.  Two    alternators   of   the   same    design   are   operating   in 
parallel.     The  first  delivers  980  kw.  at  0.95  power  factor,  and 
the  second  720  kw.  at  0.73  power  factor.     What  adjustments 
should  be  made  to  have  these  alternators  operate  under  the  best 
conditions?     When  these  have  been  made  what  power  will  each 
deliver  and  at  what  power  factor  will  it  operate? 

77.  Two  3-phase  alternators  connected  in  parallel  are  driven 
by  shunt  motors  whose  speed  load  characteristics  for  particular 
field  excitations  are  given  by  the  following  data:     The  speed  of 
the  first  motor  falls  uniformly  from  600  rev.  per  min.  at  no  load 
to  530  rev.  per  min.  at  full  load  of  100  kw.  on  the  alternator. 
The  speed  of  the  second  motor  falls  uniformly  from  590  rev.  per 
min.  at  no  load  to  550  rev.  per  min.  at  full  load  of  100  kw.  on  the 
alternator. 

(a)  For  what  load  will  the  alternators  divide  the  load  equally? 

(b)  What  will  be  the  load  on  each  alternator  when  their  com- 
bined load  is  200  kw.? 

(c)  What  is  the  greatest  load  that  can  be  delivered  without 
overloading  either  alternator? 

78.  A  3-phase,  2200-volt  alternator  which  is  rated  to  deliver 
760  kv.-a.  is  connected  in  parallel  through  transformers  with  a 
3-phase,    5500-volt   alternator   which  is  rated  to  deliver  1500 
kv.-a.     The  first  alternator  has  64  poles  and  is  driven  by  an 
engine  whose  speed  falls  from  94  rev.  per  min.  at  no  load  to 
91  rev.  per  min.  at  full  load  on  the  alternator.     The  second  al- 
ternator has  72  poles  and  is  driven  by  an  engine  whose  speed  falls 
from  83  rev.  per  min.  at  no  load  to  79  rev.  per  min.  at  full  load 
on  the  alternator. 


SYNCHRONOUS  GENERATORS  45 

(a)  What  is  the  greatest  combined  load  that  the  alternators 
can  deliver  without  overloading  either  by  more  than  25  per 
cent.? 

(b)  What  is  the  load  on  each  alternator  when  the  first  is  run- 
ning at  91.6  rev.  per  min.? 

(c)  What  is  the  frequency  when  they  are  delivering  a  com- 
bined load  of  2000  kw.? 

79.  Two  3-phase,  60-cycle  alternating-current  generators  are 
operating  in  parallel.     The  first  has  a  capacity  of  1000  and  the 
second  a  capacity  of  1500  kv.-a.     The  first  is  driven  by  a  prime 
mover  so  adjusted  that  the  frequency  falls  from  61  cycles  at  no 
load  to  59.6  cycles  at  full  load.     The  second  has  a  different  speed- 
load  characteristic,  the  frequency  falling  from  61.4  cycles  at  no 
load  to  59.2  cycles  at  full  load. 

When  these  alternators  are  jointly  delivering  2000  kw.  what  is 
the  load  on  each?  What  is  the  frequency?  If  the  speed-load 
characteristic  of  the  second  is  shifted  parallel  to  itself  until  the 
alternators  divide  this  load  properly,  what  is  the  new  value  of  the 
no-load  frequency  of  this  alternator?  At  what  frequency  will 
they  now  operate  when  delivering  2000  kw.? 

80.  Two    3-phase,    11,000-volt,    60-cycle    alternating-current 
generators,  operating  in  parallel,  are  driven  by  prime  movers 
which  have  the  same  speed-load  characteristic.     The  armature 
windings  of  the  alternators  are  Y-connected  and  have  an  effective 
resistance  of  0.94  ohm  and  a  synchronous  reactance  of  56  ohms  per 
phase.     The  total  load  supplied  is  1700  kw.  at  0.83  power  factor. 
The  excitations  are  adjusted  so  that  the  terminal  voltage  is 
11,000  volts  and  one  of  them  is  operating  at  unit  power  factor. 
What  are  the  excitation  voltages  and  the  phase  angle  between 
them? 

81.  Two  identical  3-phase,  Y-connected  alternators  operating 
in  parallel  are  driven  by  prime  movers  that  have  such  dissimilar 
speed-load    characteristics    that    when   the   first   alternator   is 
delivering  400  kw.  at  0.8  power  factor  the  second  is  running  at 
no  load,     (a)  If  the  excitations  of  the  alternators  are  adjusted  so 
that  the  terminal  voltage  is  5000  volts  and  the  armature  current 
of  the  second  alternator  is  zero,  what  are  the  excitation  voltages 
and  their  phase  displacement  ?     (b)  If  the  excitations  are  adj  usted 
so  that  the  terminal  voltage  is  5000  volts  and  the  total  armature 
copper  loss  is  reduced  to  its  least  value  what  current  will  each 
alternator  deliver?     The  effective  armature  resistance  and  the 


46     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

synchronous  reactance  of  each  alternator  are  respectively  0.42 
and  15.4  ohms  per  phase. 

82.  Two   identical  3-phase  alternators,  connected  in  parallel, 
are    driven   by  prime  movers  that  have  dissimilar  speed-load 
characteristics.     When  the  excitations   of  the   alternators   are 
equal  the  first  delivers  100  amperes  at  0.9  power  factor  (lagging) 
and  the  second,  75  amperes  at  0.7  power  factor  (lagging). 

(a)  What  per  cent,  of  the  total   load  does   each  alternator 
deliver? 

(b)  What  is  the  power  factor  of  the  load? 

(c)  If  the  field  excitations  are  adjusted  so  that  both  alter- 
nators operate  at  the  same  power  factor  what  current  will  each 
deliver? 

(d)  If  the  field  excitations  are  adjusted  so  that  the  total 
armature  copper  loss  is  reduced  to  its  least  value  at  what  power 
factor  will  each  alternator  operate? 

83.  Two  identical  2-phase  alternators,   connected  in  parallel, 
are  driven  by  prime  movers  which  have  somewhat  dissimilar 
speed-load    characteristics.     The   first  alternator  delivers  3200 
kw.  at  2210  volts  and  has  an  armature  current  of  760  amperes, 
while  the  second  delivers  3700  kw.  at  2210  volts  and  has  an 
armature   current   of   1100   amperes.     The   effective   armature 
resistance  and  the  synchronous  reactance  of  each  alternator  are 
respectively  0.0256  and  1.12  ohms  per  phase.     The  excitations 
of  the  alternators  are  now  adjusted  so  as  to  reduce  the  total 
armature  copper  loss  to  its  least  value,  but  the  terminal  voltage 
is  maintained  at  2210  volts. 

(a)  At  what  power  factor  will  each  alternator  be  operating? 

(b)  What  is  the  reduction  in  the  total  armature  copper  loss? 

(c)  What  is  the  change  in  the  copper  loss  of  each  alternator? 

84.  Two  identical  3-phase  alternators,   operating  in  parallel 
on  a  balanced  load,  are  driven  by  prime  movers  with  different 
speed-load    characteristics.     The   power   output  of  each  alter- 
nator is  measured  by  two  wattmeters.     Show  that  when  the 
differences  between  the  wattmeter  readings  for  each  alternator 
are  the  same  the  total  armature  copper  loss  is  reduced  to  its  least 
value  for  the  given  load. 

85.  Two    dissimilar    Y-connected    alternators,     operating  in 
parallel,  supply  a  load  of  1500  kw.  at  0.85  power  factor  and  a 
terminal  potential   difference   of  5000  volts.     The  alternators 
being  of  the  same  capacity  are  adjusted  to  deliver  equal  loads. 


S  YNCHRONO  US  GENERA  TORS  47 

The  effective  armature  resistances  are  respectively  0.40  and  0.52 
ohm  per  phase. 

What  should  be  the  armature  current  of  each  alternator  in 
order  that  their  combined  armature  copper  loss  will  be  reduced 
to  its  least  value? 

86.  An  alternator  which  has  a  capacity  of   1650    kv.-a.    is 
operated  in  parallel  with  one  which  has  a  capacity  of  1000  kv.-a. 
Their  armature  resistances  are  1.56  ohms  and  2.08  ohms  re- 
spectively.    For  a  combined  load  of  2200  kw.   at  0.83  power 
factor  what  load  should  each  deliver  and  at  what  power  factor 
should  it  operate  if  the  current  and  power  outputs  are  propor- 
tional to  their  ratings?     With  this  division  of  the  load  the  excita- 
tions are  adjusted  so  that  the  combined  armature  copper  loss 
is  reduced  to  a  minimum.     At  what  power  factor  should  each 
alternator  operate  for  this  latter  condition? 

87.  Two  identical  3-phase,  Y-connected  alternators  are  ope- 
rating in  parallel  with  a  common  potential  difference  of  2200 
volts.     The  effective  armature  resistance  and  the  synchronous 
reactance  of  these  alternators  are  respectively  0.158  and  2.12  ohms 
per  phase.     The  characteristics  of  the  prime  movers  are  so  dis- 
similar that  of  the  total  load  of  1500  kw.  at  0.85  power  factor 
the  first  alternator  supplies  840  kw.     The  excitations  of  the 
alternators  are  adjusted  so  that  both  are  operating  at  the  same 
power  factor.     What  are  their  excitation  voltages  and  the  phase 
displacement  between  them? 

88.  Two  identical  3-phase,  Y-connected  alternators,  operat- 
ing in  parallel,  are  driven  by  prime  movers  which  have  the  same 
speed-load    characteristic.     The    effective    armature    resistance 
and  the  synchronous  reactance  of  the  alternators  are  respec- 
tively 2.18  ohms  and  92  ohms  per  phase. 

The  alternators  supply  1830  kw.  at  13,800  volts  to  an  induction 
motor  load  that  is  operating  at  0.83  power  factor.  The  excita- 
tions of  the  alternators  are  adjusted  so  that  the  first  supplies  a 
current  of  40  amperes. 

(a)  What  current  does  the  second  alternator  supply? 

(b)  What  are  the  excitation  voltages  of  the  alternators  and 
their  phase  displacement? 

89.  Two  identical  3-phase,  Y-connected  alternators,  operating 
in  parallel,  are  driven  by  prime  movers  that  have  the  same  speed- 
load  characteristic.     The  alternators  are  rated  to  deliver  1000 
kv.-a.  at  2400  volts.     The  effective  armature  resistance  and  the 


48     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

synchronous  reactance  are  respectively  0.067  and  2.64  ohms  per 
phase.  Assume  that  the  rotational  losses,  both  core  loss  and 
friction,  are  constant  and  equal. 

(a)  If  these  alternators  are  delivering  2000  kw.  at  unit  power 
factor  and  their  rated  voltage,  by  what  amount  can  the  load  be 
shifted  from  one  to  the  other  if  their  field  excitations  are  adjusted 
so  that  there  is  an  interchange  current  equal  to  the  full-load 
current,  viz.,  240  amperes? 

(b)  If  these  alternators  are  delivering  1500  kw.  at  0.75  power 
factor  and  their  rated  voltage,  by  what  amount  can  the  load  be 
shifted  from  one  to  the  other  if  their  field  excitations  are  adjusted 
so  that  there  is  an  interchange  current  equal  to  the  full-load  cur- 
rent, viz.,  240  amperes. 

90.  The  alternators  described  in  problem  89  are  operating  in 
parallel  with  non-inductive  resistances  of  0.8  ohm  inserted  in 
each  phase  of  each  alternator. 

(a)  If  the  alternators  are  jointly  delivering  2000  kw.  at  unit 
power  factor  and  their  rated  voltage  to  a  load  by  what  amount 
can  this  load  be  shifted  from  one  to  the  other  if  the  excitations 
are  adjusted  so  that  there  is  an  interchange  current  equal  to  the 
full-load  current,  viz.,  240  amperes? 

(b)  If  the  alternators  are  jointly  delivering  1500  kw.  at  0.75 
power  factor  and  their  rated  voltage  to  a  load,  by  what  amount 
can  this  load  be  shifted  from  one  to  the  other  if  the  excitations 
are  adjusted  so  that  there  is  an  interchange  current  equal  to 
the  full-load  current,  viz.,  240  amperes? 

91.  Two  identical  3-phase  alternators,  rigidly  coupled  together 
so  that   their  excitation  voltages  are  in  phase,  are  driven  by 
a   shunt   motor.     The    effective    armature   resistance    and  the 
synchronous  reactance  of  each  alternator  are  respectively  0.302 
and  4.36  ohms  per  phase.     The  terminals  of  the  alternators  are 
electrically  connected  as  they  would  be  for  parallel  operation  but 
no  external  load  is  supplied.     When  the  field  excitations  are  ad- 
justed so  that  the  excitation  voltages  are  respectively  200  and 
300  volts  per  phase,   what  is  the  armature   current?     If  the 
armature  windings  are  connected  in  delta  what  is  the  terminal 
voltage?     What    is    the    electrical    output    of    the    alternator 
which    is   acting  as  a  generator?     If  the  rotational  losses  are 
supplied  by  the  shunt  motor  what  is  the  mechanical  output  of 
the  alternator  which  is  acting  as  a  motor?     If  the  rotational 
losses  are  1620  watts  what  power  does  the  shunt  motor  supply? 


SYNCHRONOUS  GENERATORS  49 

92.  If  the  alternators  described  in  problem  91  are  mechanically 
coupled  together  so  that  their  excitation  voltages  differ  in  phase 
by  30  degrees  what  will  be  the  armature  current  when  the 
excitation  voltages  are  each  300  volts?     What  is  their  terminal 
voltage?     What  is  the  electrical  output  of  the  alternator  which  is 
acting  as  a  generator?     If  the  rotational  losses  are  supplied  by 
the  shunt  motor  what  is  the  mechanical  output  of  the  alternator 
which  is  acting  as  a  motor?     What  power  does  the  shunt  motor 
supply  if  the  rotational  losses  are  1620  watts? 

93.  Two    identical,    Y-connected,    60-cycle    alternators    are 
rigidly  coupled  together  and  are  driven  at  their  rated  speed  of 
1200  rev.  per  min.     The  alternators  have  revolving  fields  and 
the  coupling  is  so  made  that  the  north  poles  of  the  first  are  10 
degrees  (mechanical)  ahead,  i.e.,  in  the  direction  of  rotation,  of 
the  corresponding  north  poles  of  the  second.     The  corresponding 
terminals   are   connected   through  non-inductive  resistances   of 
1.5    ohms    each.     The   effective   armature   resistance    and   the 
synchronous  reactance  of  each  alternator  are  respectively  0.302 
and  4.36  ohms  per  phase.     The  field  currents  are  adjusted  so 
that  the  excitation  voltages  are  respectively  200  and  300  volts 
per  phase.     What  is  the  current?     What  is  the  electrical  out- 
put of  the  alternator  which  is  acting  as  a  generator?     If  the 
rotational  losses  are  supplied  by  the  driving  motor  what  is  the 
mechanical  output  of  the  alternator  which  is  acting  as  a  motor? 
If  the  rotational  losses  are  1620  watts  what  is  the  output  of  the 
driving  motor? 

94.  Two  identical,  3-phase,  Y-connected  alternators,  rigidly 
coupled  to  the  same  prime  mover,  are  operating  in  parallel  and 
supply  1500  kw.  at  0.83  power  factor  and  a  terminal  potential 
difference  of  5000  volts.     The  effective  armature  resistance  and 
the  synchronous  reactance  of  each  alternator  are  respectively 
0.42  and  15.4  ohms  per  phase.     The  mechanical  coupling  is  so 
made  that  the  excitation  voltages  are  in  phase,  and  the  field 
currents  are  adjusted  so  that  one  of  these  voltages  is  50  per  cent, 
greater  than  the  other.     What  is  the  output  of  each  alternator? 

95.  Two  3-phase,  Y-connected  alternators,  each  of  which  is 
rated  to  deliver  760  kv.-a.  at  2200  volts,  are  rigidly  coupled  to  the 
same  prime  mover.     Each  of  these  alternators  has  64  field  poles, 
an  effective  armature  resistance  of  0.172  ohm  and  a  synchronous 
reactance  of  2.12  ohms  per  phase.     With  equal  field  excitations 
they  operate  at  full  load  with  undue  heating  and  on  examination 


50     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

it  is  found  that  their  corresponding  field  poles  are  displaced  by 
an  angle  of  1  degree  and  6  minutes. 

(a)  When  they  deliver  1500  kw.  at  0.87  power  factor  and 
their  rated  voltage,  what  is  the  output  of  each  alternator  if  the 
field  excitations  are  equal? 

(b)  If  this  displacement  of  the  field  poles  is  reduced  to  zero 
what  will  be  the  reduction  in  the  armature  copper  loss  of  each 
alternator?     The  load  delivered  is  still  1500  kw.  at  0.87  power 
factor  and  their  rated  voltage. 

96.  Two  identical,  2-phase,  3750-kv.-a.   water-wheel  genera- 
tors, operating  in  parallel,  are  driven  by  prime  movers  which  have 
the  same  speed-load  characteristic.     The  effective  armature  resis- 
tance and  the  synchronous  reactance  of  these  alternators  are 
respectively  0.0254  and  1.12  ohms  per  phase.     The  alternators 
are  jointly  delivering  7500  kw.  at  unit  power  factor  and  their 
rated  terminal  potential  difference  of  2200  volts.     The  excita- 
tions are  equal. 

Instantaneous  records  show  that  due  to  hunting  the  maximum 
displacement  of  their  excitation  voltages  is  30  degrees.  What  is 
the  synchronizing  power  at  the  time  of  maximum  displace- 
ment? What  is  the  maximum  value  of  the  effective  armature 
current? 

97.  If  the  alternators  described  in  problem  96  are  jointly 
delivering  5250  kw.  at  0.7  power  factor  and  their  rated  voltage 
to  an  inductive  load,  what  is  the  synchronizing  power  at  the 
time  the  displacement  of  their  excitation  voltages  is  30  degrees? 
What  are  the  effective  armature  currents  at  this  time? 

98.  If  the  alternators   described  in  problem  96  are  jointly 
delivering  5250  kw.  at  0.7  power  factor  and  their  rated  voltage 
to  a  condensive  load,  what  is  the  synchronizing  power  at  the 
time  the  displacement  of  their  excitation  voltages  is  30  degrees? 
What  are  the  effective  armature  currents  at  this  time? 

99.  Two  identical  3-phase,  Y-connected  alternators,  each  of 
which  is  rated  to  deliver  1640  kv.-a.  at  13,500  volts  are  operating 
in  parallel.     The  speed-load  characteristics  of  the  prime  movers 
are  the  same  and  the  load  requires  70  amperes  from  each  alter- 
nator at  0.83  power  factor  and  the  rated  voltage.     The  effective 
armature  resistance  and  the  synchronous  reactance  of  the  alter- 
nators are  respectively  1.52  ohms  and  37.4  ohms  per  phase. 
Oscillograph  records   show  that   due  to  hunting  the  greatest 
value  of  the  effective  armature  current  is  87  amperes. 


SYNCHRONOUS  GENERATORS  51 

What  is  the  maximum  displacement  between  the  excitation 
voltages? 

What  is  the  synchronizing  power  at  this  maximum  displace- 
ment? 


CHAPTER  III 
SYNCHRONOUS  MOTORS 

1.  A  3-phase,  60-cycle,  A-connected  alternating-current 
generator  is  rated  to  deliver  15  kv.-a.  at  230  volts.  The  field 
structure  has  6  poles  each  of  which  is  wound  with  398  turns. 
The  armature  core  has  4  slots  per  pole  per  phase  with  8  inductors 
in  series  per  slot.  Each  inductor  is  5  in.  long.  The  hot  resist- 
ance of  the  armature  measured  between  any  two  terminals  is 
0.126  ohm.  The  effective  resistance  is  1.6  times  the  ohmic 
resistance.  In  calculating  the  leakage  reactance  assume  8 
leakage  lines  per  ampere  per  inch  of  inductor.  The  open-  and 
short-circuit  characteristics  are  given  by  the  following  data: 


Open  circuit 

Short  circuit 

Terminal  voltage 

Armature  current 

2.0 
4.0 
5.0 
6  0 

104 
199 
240 
275 

22 

44 
55 

7.5 

315 

This  generator  is  running  as  an  overexcited  synchronous  motor 
and  receives  15  kw.  at  its  rated  voltage  and  0.8  power  factor. 

(a)  Calculate  the  field  current  by  the  general  method. 

(b)  Calculate  the  field  current  by  the  synchronous-impedance 
method. 

(c)  Calculate  the  field  current  by  the  magnetomotive-force 
method. 

2.  A  3-phase,  5000-volt,  25-cycle  synchronous  motor  has  a 
full-load  capacity  of  1100  h.p.  The  field  structure  consists  of  32 
poles  each  of  which  is  wound  with  79.5  turns.  The  armature 
core  has  192  slots  with  14  inductors  in  series  per  slot.  The 
armature  winding  is  Y-connected.  The  effective  resistance  of 
the  armature  winding  is  0.38  ohm  per  phase.  In  calculating  the 
leakage  reactance,  assume  75  leakage  lines  per  ampere  per 
inductor.  The  open-  and  short-circuit  characteristics  are  given 
by  the  following  data: 

52 


SYNCHRONOUS  MOTORS 


53 


Field  ampere- 

Open  circuit 

Short  circuit 

turns  per  pole 

Terminal  voltage 

Armature  current 

4,000 
6,000 
10,000 

3200 
4330 
5460 

108 
162 

12,000 

5800 

14,000 

6060 

This  motor  receives  760  kw.  at  its  rated  voltage  and  0.83  power 
factor.  The  excitation  is  less  than  normal. 

(a)  Calculate  the  field  current  by  the  general  method. 

(b)  Calculate  the  field  current  by  the  synchronous-impedance 
method. 

(c)  Calculate  the  field  current  by   the  magnetomotive-force 
method. 

3.  A  3-phase,  5500-volt,  50-cycle  synchronous  motor  has  a 
full-load  capacity  of  2000  h.p.  The  field  structure  has  72 
poles  each  of  which  is  wound  with  35  turns.  The  armature  core 
has  one  slot  per  pole  per  phase  with  12  inductors  per  slot.  The 
armature  winding  is  connected  in  Y.  The  armature  resistance 
measured  between  any  two  terminals  is  0.536  ohm,  and  the  ratio 
of  effective  to  ohmic  resistance  is  1.35.  In  calculating  the 
leakage  reactance  assume  67  leakage  lines  per  inductor  per 
ampere.  The  data  for  the  open-  and  short-circuit  characteristics 
are: 


Open  circuit 

Short  circuit 

Field  current 

Terminal  voltage 

Armature  current 

150 

200 
250 

5100 
5900 
6500 

300 

400 

300 

6800 

350 

7100 

This  motor  receives  960  kw.  at  its  rated  voltage  and  0.68  power 
factor.     The  excitation  is  greater  than  normal. 

(a)  Calculate  the  field  current  by  the  general  method. 

(b)  Calculate  the  field  current  by  the  synchronous-impedance 
method. 

(c)  Calculate  the  field  current  by  the  magnetomotive-force 
method. 

4.  A  3-phase,  2200-volt,  50-cycle  synchronous  motor  is  rated 
to  deliver  1000  h.p.  when  operating  at  unit  power  factor.    The 


54     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 


field  structure  has  64  poles  each  of  which  is  wound  with  50  turns. 
The  resistance  of  the  field  circuit  is  0.516  ohm.  The  armature 
core  has  384  slots  with  3  inductors  in  series  per  slot.  Each  in- 
ductor is  10  in.  long.  The  hot  resistance  of  the  armature  meas- 
ured between  terminals  is  0.264  ohm.  The  ratio  of  effective 
to  ohmic  resistance  is  1.3.  In  calculating  the  leakage  reactance 
assume  6.5  lines  per  ampere  per  inch  of  inductor.  The  armature 
windings  are  connected  in  Y.  The  open-  and  short-circuit 
characteristics  are  given  by  the  following  data: 


Open  circuit 

Short  circuit 

Terminal  voltage 

Armature  current 

40 
50 
60 

1450 
1840 
2220 

225 

280 

80 

2860 

100 

3380 

The  rotational  losses  at  normal  voltage  are  17.2  kw. 

This  motor  deliveres  1000  h.p.  and  the  excitation  greater  than 
normal  and  is  adjusted  so  that  it  is  operating  from  a  2200-volt 
circuit  at  a  power  factor  of  0.93. 

(a)  Calculate  the  voltage  impressed  on  the  field  circuit     Use 
the  general  method. 

(b)  Calculate  the  voltage  impressed  on  the  field  circuit.     Use 
the  synchronous-impedance  method. 

(c)  Calculate  the  voltage  impressed  on  the  field  circuit.     Use 
the  magnetomotive-force  method. 

5.  A  2-phase,  10,000-volt  synchronous  motor  is  rated  to  deliver 
4500  h.p.  when  operating  at  unit  power  factor.  The  armature 
has  an  effective  resistance  of  0.64  ohm  per  phase.  The  open- 
and  short-circuit  characteristics  are  given  by  the  following  data: 


Open  circuit 


Short  circuit 


rieia  current 

Terminal  voltage 

Armature  current 

100 
200 
250 

7,700 
10,200 
10,900 

220 
430 

300 

11,500 

375 

12,200 

This  motor  is  delivering  3760  h.p.  with  an  efficiency,  exclusive 
of  field  loss,  of  95.6  per  cent.,  and  the  field  current  is  adjusted 
so  that  it  takes  a  leading  current  at  0.91  power  factor. 


SYNCHRONOUS  MOTORS  55 

(a)  Calculate  the  field  current  by  the  synchronous-impedance 
method. 

(b)  Calculate  the  field  current  by  the  magnetomotive-force 
method. 

6.  The  full-load  capacity  of  a  3-phase,  13,500-volt,  synchro- 
nous motor  is  2200  h.p.  The  armature  windings,  which  are 
connected  in  Y,  have  an  effective  resistance  of  1.52  ohm  per 
phase.  The  open-  and  short-circuit  characteristics  are  given  by 
the  following  data: 


Open  circuit 

Short  circuit 

Field  current 

Terminal  voltage 

Armature  current 

50 
100 
150 
200 

7,500 
10,100 
14,700 
15,800 

75 
155 
227 

250 

16,700 

Calculate  the  field  current  by  the  synchronous  impedance 
method  when  this  motor  receives  1640  kv.-a.  at  0.75  power 
factor,  (a)  with  excitation  greater  than  normal,  (b)  with  excita- 
tion less  than  normal. 

7.  The  synchronous  motor  described  in  problem  6  receives 
1250  kw.  at  0.75  power  factor. 

Calculate  the  field  current  by  the  magnetomotive-force 
method  (a)  with  excitation  greater  than  normal,  (b)  with  excita- 
tion less  than  normal. 

8.  Concerning  the  alternator  described  in  problem  1  the  fol- 
lowing additional  data  are  given:  The  field  current  is  supplied 
at  110  volts.     The  friction  and  windage  loss  is  310  watts  at 
normal  speed.     The  core  loss  is  280  watts,  480  watts,  and  610 
watts  for  generated  armature  voltages  of  199  volts,  240  volts, 
and  275  volts  respectively. 

What  is  the  efficiency  of  this  generator  when  it  is  running  as 
an  overexcited  synchronous  motor  and  receives  15  kw.  at  0.8 
power  factor? 

Calculate  the  field  current  (a)  by  the  general  method,  (b)  by 
the  synchronous-impedance  method,  and  (c)  by  the  magneto- 
motive-force method. 

9.  Concerning  the  synchronous  motor  described  in  problem  2 
the  following  additional  data  are  given:  The  resistance  of  the 
field  circuit  is  0.82  ohm.     The  friction  and  windage  loss  at 


56      PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

normal  speed  is  7.6  kw.  The  core  loss  due  to  rotation  is  22.3 
kw.  at  5460  volts  and  may  be  assumed  to  vary  as  the  square  of 
the  generated  armature  voltage.  What  is  the  efficiency  of 
this  motor  under  the  conditions  described  in  problem  2? 

Calculate  the  field  current  (a)  by  the  general  method,  (b)  by 
the  synchronous-impedance  method,  and  (c)  by  the  magneto- 
motive-force method. 

10.  Concerning  the  synchronous  motor  described  in  problem 
3  the  following  additional  data  are  given.     The  field  circuit  has 
a  resistance  of  0.376  ohm.     The  friction  and  windage  loss  is  8.4 
kw.     The  core  loss  due  to  rotation  is  20.2  kw.  at  5500  volts  and 
may  be  assumed  to  vary  as  the  square  of  the  generated  armature 
voltage.     What  is  the  efficiency  of  this  motor  under  the  conditions 
described  in  problem  3  ? 

Calculate  the  field  current  (a)  by  general  method,  (b)  by  the 
synchronous-impedance  method,  and  (c)  by  the  magnetomotive- 
force  method. 

11.  A    3-phase,    2200-volt    synchronous    motor    is    rated    to 
deliver  1000  h.p.  when  operating  at  unit  power  factor.     The 
armature  windings,  which  are  connected  in  Y,  have  an  effective 
resistance  of  0.172  ohm  per  phase.     The  resistance  of  the  field 
circuit  is  0.576  ohm.     The  friction,  windage  and  core  losses  are 
17.2  kw.  and  may  be  assumed  constant.     The  open-  and  short- 
circuit  characteristics  are  given  by  the  following  data: 


. 

Open  circuit 

Short  circuit 

Terminal  voltage 

Armature  current 

40 
•     50 
60 

80 

1450 
1840 
2220 
2860 

225 

280 

100 

3380 

What  is  the  efficiency  of  this  motor  when  it  delivers  960  h.p.  and 
is  operating  at  0.83  power  factor  from  a  2200-volt  circuit  (a)  if 
the  excitation  is  greater  than  normal,  (b)  if  the  excitation  is  less 
than  normal? 

Calculate  the  field  current  by  the  synchronous  impedance 
method. 

12.  A  2-phase,  2200-volt  synchronous  motor  has  a  full-load 
capacity  of  5000  h.p.  when  operating  at  unit  power  factor.  The 
armature  has  an  effective  resistance  of  0.0255  ohm  per  phase. 


SYNCHRONOUS  MOTORS 


57 


The  resistance  of  the  field  circuit  is  0.37  ohm.  The  friction  and 
windage  loss  is  19.0  kw.  The  core  loss  due  to  rotation  is  45.2 
kw.  at  2200  volts  and  may  be  assumed  constant.  The  open- 
and  short-circuit  characteristics  are  given  by  the  following  data: 


Field  current 

Open  circuit 

Short  circuit 

Terminal  voltage 

Armature  current 

66.7 
133.0 
200  0 

680 
1360 
2000 

606 
1210 

267  0 

2480 

300.0 

2660 

What  is  the  efficiency  of  this  motor  when  it  receives  3200  kw. 
at  0.80  power  factor  (a)  if  the  excitation  is  greater  than  normal, 
(b)  if  the  excitation  is  less  than  normal? 

Calculate  the  field  current  by  the  magnetomotive-force 
method. 

13.  The  following  test  data  are  given  on  a  1340-h.p.,  11,000- 
volt,  3-phase  synchronous  motor.  The  armature  effective  resist- 
ance is  0.94  ohm  per  phase,  and  the  resistance  of  the  field  winding 
is  3.11  ohms.  The  armature  windings  are  connected  in  Y. 


Field 
current 

Open  circuit  ter- 
minal voltage  (  V) 

Core  loss 
and  V 

Terminal  voltage 
for/0  =  60,  PF  =  Q 

20 

8,400 

11,600 

30 
40 
50 

11,000 
12,700 
13  800 

19,400 
25,600 

9  800 

60 
70 

14,700 
15.500 

11,100 
12.100 

The  friction  and  windage  loss  is  12.1  h.p. 

What  is  the  efficiency  of  this  motor  when  it  receives  68  amperes 
per  terminal  at  11,550  volts  if  the  excitation  is  greater  than 
normal  and  is  adjusted  so  that  the  power  factor  is  78.4  per  cent.? 

14.  The  following  test  data  are  given  on  a  6500-h.p.,  6600- 
volt,  3-phase  synchronous  motor.  The  effective  resistance  of 
the  armature  winding  is  0.081  ohm  per  phase,  and  the  resistance 
of  the  field  winding  is  0.549  ohm.  The  armature  windings  are 
connected  in  Y. 


58     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 


Field 
current 

Open  circuit  ter- 
minal voltage  (V) 

Core  loss 
and  V 

Terminal  voltage 
7a  =  438,  PF=0 

100 
150 

200 
250 

4800 
6470 
7400 
7930 

45  kw. 
85  kw. 
120  kw. 

1750 
4200 
5700 
6600 

275 

8150 

6900 

The  friction  and  windage  loss  is  62.8  h.p. 

What  is  the  efficiency  of  this  motor  when  it  receives  450 
amperes  at  unit  power  factor,  and  its  rated  voltage? 

15.  The  following  test  data  are  given  on  a  1340-h.p.,  13,800- 
volt,  3-phase  synchronous  motor.  The  armature  windings, 
which  are  connected  in  Y,  have  an  effective  resistance  of  2.18 
ohms  per  phase.  The  resistance  of  the  field  circuit  is  0.541  ohm. 


Field 
current 

Open  circuit  ter- 
minal voltage  (V) 

Terminal  voltage 
la  =  42,  PF  =  0 

Core  loss 
and  V 

50 

8,800 

7  5 

80 
110 
140 
180 

13,000 
15,600 
17,250 

18.900 

. 

10,750 
13,250 
15.600 

16.6 
25.4 
33.5 

The  friction  and  windage  loss  is  11.2  h.p. 

What  is  the  efficiency  of  this  motor  when  it  delivers  1200  h.p. 
if  the  excitation  is  greater  than  normal  and  is  adj  usted  so  that 
the  motor  takes  an  armature  current  of  47.6  amperes  at  a 
terminal  voltage  of  13,500  volts? 

16.  A    synchronous    motor,    whose    armature    windings    are 
connected  in  Y,  has  an  effective  resistance  of  0.94  ohm  and  a 
synchronous   reactance    of   56    ohms   per   phase.     This   motor 
receives  a  line  current  of  60  amperes  at  a  terminal  potential 
difference  of  11,000  volts  and  the  excitation  is  adjusted  so  that 
the  power  factor  is  0.85. 

What  power  does  the  motor  receive?  What  is  the  exci- 
tation voltage  (a)  if  the  excitation  is  greater  than  normal, 
(b)  if  the  excitation  is  less  than  normal? 

17.  A  2-phase  synchronous  motor  has  an  effective  armature  re- 
sistance of  0.64  ohm  and  a  synchronous  reactance  of  23.7  ohms 
per  phase.     The  motor  receives  3000  kw.  at  a  line  potential 
difference  of   10,000  volts  and  the  excitation  adjusted  so  that 
the  line  current  is  200  amperes. 


SYNCHRONOUS  MOTORS  59* 

At  what  power  factor  is  this  motor  operating?  What  is  the 
excitation  voltage  (a)  if  the  excitation  is  greater  than  normal, 
(b)  if  the  excitation  is  less  than  normal? 

18.  A  5000-h.p.  synchronous  motor  is  operated  from  a  2200- 
volt,  2-phase  circuit.  The  effective  resistance  of  the  armature 
is  0.0254  ohm  per  phase.  The  field  winding  has  60  turns  per  pole 
and  a  total  resistance  of  0.37  ohm. 


Ampere  turns  per 
pole 

Open  circuit  terminal 
voltage 

Short  circuit  armature 
current 

4,000 
8,000 
12,000 

680 
1360 
2000 

606 
1210 

16,000 
18,000 

2480 
2660 

At  no  load  this  motor  takes  67.5  kw.  at  2200  volts  when  the 
excitation  is  adjusted  so  that  it  operates  at  0.86  power  factor. 
How  much  must  the  excitation  be  increased  in  order  that  the 
motor  will  deliver  its  rated  load  and  operate  at  this  same  power 
factor?  Use  the  magnetomotive-force  method  for  calculating 
the  field  current. 

19.  A  3-phase  synchronous  motor,  whose  armature  windings 
are  connected  in  A,  has  an  effective  resistance  of  0.302  ohm  and 
a  synchronous  reactance  of  4.36  ohms  per  phase.     This  motor 
receives  a  line  current  of  40  amperes  at  230  volts  and  the  field 
current  is  adjusted  so  that  the  excitation  voltage  is  310  volts. 

What  power  does  the  motor  receive?  At  what  power  factor 
is  it  operating? 

20.  A  2400-volt,  3-phase,  synchronous  motor  has  a  full-load 
capacity  of  1340  h.p.     The  effective  resistance  of  the  armature 
is  0.067  ohm  per  phase  and  the  synchronous  reactance  is  2.64 
ohms  per  phase.     The  armature  windings  are  connected  in  Y. 
The  resistance  of  the  field  circuit  is  0.427  ohm.     The  rotational 
losses  at  normal  voltage  are  26.6  kw. 

This  motor  is  operated  from  a  2400-volt  circuit  at  0.82  power 
factor  and  is  delivering  1220  h.p.  What  is  the  line  current, 
and  what  is  the  necessary  excitation  voltage?  Calculate  the 
excitation  voltage  for  both  under-  and  overexcitation. 

21.  A  135-h.p.  synchronous  motor  is  operated  from  a  480-volt, 
2-phase   circuit.     The  effective  resistance  of  the  armature  is 
0.135   ohm  per  phase  at  normal  running  temperature.     The 
rotational  losses  at  normal  voltage  are  5.7  kw. 


60     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 


Field  current         Terminal  voltage  on  open  ;      Armature  current  on 
circuit  short  circuit 


10 
17 

397 
527 

158 

25 

600 

35 

655 

With  the  greatest  allowable  excitation  the  motor  takes  a 
current  of  135  amperes  when  delivering  its  rated  load.  At  what 
power  factor  does  it  then  operate?  What  is  the  field  current? 
Use  what  you  consider  the  best  method  for  calculating  the  field 
current. 

22.  A  3-phase  synchronous  motor,  whose  armature  windings  are 
connected  in  Y,  has  an  effective  resistance  of  0.172  ohm  and  a 
synchronous  reactance  of  2.12  ohms  per  phase.     This  motor 
receives  750  kw.  at  a  line  potential  difference  of  2200  volts, 
and  the   field  current  is  adjusted  so  that  the  excitation  voltage 
is  2800  volts. 

What  is  the  line  current?  At  what  power  factor  is  the  motor 
operating? 

23.  A  2200-volt,  2-phase  synchronous  motor  has  a  full-load 
capacity  of  5000  h.p.     The  full-load  efficiency  of  the  armature, 
with  the  field  current  adjusted  for  unit  power  factor,  is  96.1  per 
cent.     The  armature  winding    has    an    effective  resistance    of 
0.025  ohm  and  a  synchronous  reactance  of  1.12  ohms  per  phase. 
The  motor  receives  a  constant  power  of  3000  kw.  at  2200  volts. 
If  the  current  is  limited  to  130  per  cent,  of  its  full-load  value, 
what  is  the  greatest  allowable  excitation  voltage?     At  what 
power  factor  would  the  motor  be  operating? 

24.  A  3-phase,    13,500-volt  synchronous   motor  is  rated   to 
deliver  2200  h.p.  when  operating  at  unit  power  factor.     The 
armature  has  an  effective  resistance  of  1.52  ohms  per  phase. 
The  armature  windings  are  connected  in  Y. 


„.  . .  Open  circuit  terminal          Short  circuit  armature 

Field  current 

voltage  current 


50 

7,500 

75 

100 

10,100 

155 

150 

14,700 

227 

The  rotational  losses  at  normal  voltage  are  68  h.p. 

If  the  maximum  allowable  current  is  125  per  cent,  of  the  full- 


SYNCHRONOUS  MOTORS  61 

load  current  over  what  range  should  it  be  possible  to  vary  the 
field  current  when  the  motor  is  delivering  a  constant  load  of 
2000  h.p.?  Use  the  magnetomotive-force  method  for  calculat- 
ing the  field  current. 

25.  A  3-phase,  230-volt,  synchronous  motor,  whose  armature 
windings  are  connected  in  A,  has  an  effective  resistance  of  0.302 
ohm  and  a  synchronous  reactance  of  4.36  ohms  per  phase.     The 
rotational  losses  are  750  watts  and  may  be  assumed  constant. 
The  motor  delivers  a  constant  load  of  20  h.p.     What  is  the 
least  excitation  voltage  with  which  the  motor  will  run?     What 
is  the  armature  current  at  the  instant  of  breakdown? 

26.  A  1340-h.p.,  3-phase  synchronous  motor  receives  a  con- 
stant power  of  860  kw.  from  an  11,000-volt  circuit.     The  arma- 
ture windings,   which  are  connected  in  Y,   have  an  effective 
resistance  of  0.94  ohm  and  a  synchronous  reactance  of  56  ohms 
per  phase. 

Over  what  range  can  the  excitation  voltage  be  varied  so  that 
the  current  will  not  exceed  135  per  cent,  of  its  full-load  value, 
which  is  62.5  amperes? 

27.  A  3-phase,  5000-volt  synchronous  motor,  whose  armature 
windings  are  connected  in  Y,  has  an  effective  resistance  of  0.40 
ohm  and  a  synchronous  reactance  of  15.4  ohms  per  phase.     The 
rotational  losses  are  30  kw.  and  may  be  assumed  constant.     The 
greatest  excitation  voltage  that  can  be  obtained  is  3520  volts  per 
phase. 

When  the  motor  is  delivering  its  full  load  of  1100  h.p.  over 
what  range  can  the  power  factor  be  varied?  What  is  the  arma- 
ture current  at  each  of  the  limiting  conditions?  Compare  these 
with  the  full-load  armature  current. 

28.  Neglecting  the  field  copper  loss,  the  efficiency  of  a  2200- 
h.p.,  3-phase  synchronous  motor  is  96.0  per  cent,  at  full  load 
when  the  motor  is  operating  at  unit  power  factor.     Assume  that 
the  rotational   losses   are   constant.     The   armature   windings, 
which  are  connected  in  Y,  have  an  effective  resistance  of  1.52 
ohms  and  a  synchronous  reactance  of  37.4  ohms  per  phase. 
With  the  motor  delivering  2000  h.p.  the  field  current  is  adjusted 
so  that  the  motor  takes  a  leading  current  of  85  amperes  from  a 
13,500-volt  constant  potential  circuit. 

(a)  At  what  power  factor  is  the  motor  operating?     What  is 
the  excitation  voltage? 

(b)  If  this  load  is  thrown  off  what  current  will  the  motor  take, 


62     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

and  at  what  power  factor  will  it  be  operating?     The  excitation 
is  unchanged. 

29.  The  rotational  losses  of  a  1340-h.p.,  2400-volt,  3-phase 
synchronous  motor  are  43  h.p.,  and  may  be  assumed  constant. 
The  armature  windings,  which  are  connected  in  Y,  have  an 
effective  resistance  of  0.067  ohm  and  a  synchronous  reactance 
of  2.64  ohms  per  phase. 

(a)  What  is  the  least  power  factor  at  which  the  motor  can  be 
operated  at  no  load  so  that  the  current  will  not  exceed  135  per 
cent,  of  its  full-load  value?  If  the  motor  is  overexcited  what  is 
the  excitation  voltage? 

30.  When  operating  at  unit  power  factor  the  full-load  losses 
of    a    1100-h.p.,    5000-volt,    3-phase    synchronous    motor    are: 
Armature  copper  loss=13.9  kw:  Field  copper  loss=18.6  kw: 
Rotational  losses  =  27. 8  kw.     Assume  that  the  rotational  losses 
are  constant.     The  armature  windings,  which  are  connected  in 
Y,  have  a  synchronous  impedance  of  15.4  ohms  per  phase. 

If  the  current  is  limited  to  130  per  cent,  of  its  full-load  value 
what  is  the  least  power  factor  at  which  this  motor  can  be  oper- 
ated when  it  is  delivering  full  load?  What  is  the  necessary 
excitation  voltage  if  the  motor  is  overexcited? 

31.  The  synchronous  motor  described  in  problem  14  is  rated 
to  deliver  6500  h.p.  when  operating  at  unit  power  factor.     If 
the  maximum  allowable  current  is  130  per  cent,  of  its  full-load 
value  what  is  the  least  power  factor  at  which  it  can  operate  when 
delivering  its  rated  load?     What  is  the  greatest  allowable  value 
of  the  field  current  at  full  load  under  this  condition? 

32.  A  20-h.p.,   230-volt,   3-phase  synchronous  motor  has  an 
effective  resistance  of  0.302  ohm  and  a  synchronous  reactance  of 
4.36  ohms  per  phase.     The  armature  windings  are  connected 
in  A.     The  rotational  losses  are  750  watts  and  may  be  assumed 
constant. 

With  the  maximum  excitation  voltage  of  315  volts  at  what 
load  will  this  motor  break  down?  Compare  the  current  at 
breakdown  with  the  full-load  current  for  normal  excitation. 

33.  If  the  ratio   of  resistance   to   synchronous  reactance  is 
increased  to  0.30  by  inserting  equal  resistances  in  series  with 
each  phase  of  the  synchronous  motor  what  will  be  the  results 
called  for  (a)  in  problem  25;  (b)  in  problem  26;  (c)  in  problem 
27;  (d)  in  problem  29;  (e)  in  problem  30;  (f)  in  problem  32. 

34.  A  3-phase  synchronous  motor  receives  line  currents  of 


SYNCHRONOUS  MOTORS  63 

226  amperes  at  a  terminal  potential  difference  of  2180  volts  and 
delivers  812  h.p.  The  no-load  rotational  losses  are  23  h.p.,  and 
may  be  assumed  constant.  The  armature  windings,  which  are 
Y-connected,  have  an  effective  resistance  of  0.172  ohm  and  a 
synchronous  reactance  of  2.12  ohms  per  phase. 

At  what  power  factor  is  the  motor  operating?  Neglecting  the 
field  copper  losses,  what  is  the  efficiency? 

35.  At  no  load  a  3-phase  synchronous  motor  takes  line  currents 
of  1.52  amperes  at  11,000  volts  if  the  field  current  is  adjusted  for 
a  minimum  armature  current.     The  armature  windings,  which 
are  connected  in  Y,  have  an  effective  resistance  of  0.94  ohm  and 
a  synchronous  reactance  of  56  ohms. 

When  the  motor  receives  960  kw.  at  11,000  volts  and  the 
excitation  is  adjusted  so  that  the  line  current  is  72  amperes,  what 
is  the  output?  If  the  excitation  is  greater  than  normal,  what  is 
the  excitation  voltage? 

36.  The  no-load  rotational  losses  of  a  1340-h.p.,  3-phase  syn- 
chronous motor  are  34.6  kw.     The  armature  windings  are  con- 
nected in  Y  and  have  an  effective  resistance  of  0.067  ohm  and  a 
synchronous  reactance  of  2.64  ohms  per  phase.     This  motor  is 
operated  from  a  2400-volt  circuit  and  when  delivering  its  rated 
output  the  excitation  is  adjusted  so  that  the  motor  takes  a  leading 
current  of  0.83  power  factor. 

What  is  the  power  input?     What  is  the  excitation  voltage? 

37.  At  no  load  a  150-h.p.,  2-phase  synchronous  motor  takes 
7.6  kw.  at  480  volts  when  the  excitation  is  adjusted  for  a  minimum 
armature  current.     Under  load  this  motor  takes  108  amperes  at 
485  volts  when  the  excitation  is  adjusted  so  that  the  power  factor 
is  0.78.     The  effective  resistance  of  the  armature  is  0.118  ohm 
and  the  synchronous  reactance  is  2.13  ohms. 

What  is  the  output?  What  is  the  excitation  voltage  (a)  if 
the  field  current  is  greater  than  normal?  (b)  if  the  field  current 
is  less  than  normal? 

38.  A  3-phase,  6600-volt  synchronous  motor  is  operated  at 
its  rated  voltage  and  delivers  its  rated  output  of  6500  h.p. 
In  order  that  the  motor  shall  take  a  large  leading  current  the 
field  current  is  adjusted  for  an  excitation  voltage  of  8100  volts. 
The  no-load  rotational  losses  are  138  kw.  and  may  be  assumed 
constant.     The  armature  windings  are  connected  in  Y  and  have 
an  effective  resistance  of  0.081  ohm  and  a  synchronous  reactance 
of  2.8  ohms. 


64      PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

What  is  the  power  factor?  What  current  does  the  motor 
take? 

39.  A  2-phase,   10,000-volt  synchronous  motor  has  a  rated 
capacity  of  4500  h.p.     The  armature  has  an  effective  resistance 
of  0.64  ohm  and  a  synchronous  reactance  of  23.7  ohms  per  phase. 
At  no  load  with  an  excitation  greater  than  normal  the  motor 
takes  74  kw.  at  10,000  volts  and  the  line  current  is  41  amperes. 

With  the  field  excitation  unchanged  what  current  will  the 
motor  take  when  it  is  delivering  its  rated  output?  At  what 
power  factor  will  the  motor  be  operating? 

40.  At  the  end  of  a  3-phase  transmission  line  are  induction 
motors  which  take  a  total  load  of  1200  kw.  at  0.87  power  factor. 
An  additional  load  of  600  h.p.  should  be  provided  for.     Find  the 
kilovolt-ampere    capacity    of   a    synchronous   motor    that   will 
supply  this  load  and  will  at  the  same  time  make  it  possible  to 
adjust  the  power  factor  of  the  entire  load  to  unity.     Assume  that 
the  efficiency  of  the  synchronous  motor  is  92  per  cent. 

41.  An  induction  motor  load  at  the  end  of  a  3-phase  trans- 
mission line  takes  6000  kw.  at  0.90  power  factor.     A  synchronous 
motor  is  operating  in  parallel  with  the  induction  motors  in  order 
to  improve  the  power  factor.     The  motor  has  a  full-load  capacity 
of  6500  h.p.  when  operating  at  unit  power  factor  from  a  6600-volt 
circuit.     The  armature  windings,  which  are  connected  in  Y,  have 
an  effective  resistance  of  0.081  ohm  per  phase.     The  synchronous 
impedance  is  3.57  ohms  per  phase.     The  friction  and  windage 
losses  are  62.8  h.p.  and  the  core  losses  at  normal  voltage  are  86 
kw.     The  latter  may  be  assumed  to  be  constant.     The  synchro- 
nous motor  is  operated  so  that  the  resultant  power  factor  of  the 
load  is  unity,  and  the  line  voltage  is  6600  volts.     If  the  current 
is  limited  to  125  per  cent,  of  its  full-load  value  what  is  the 
greatest  power  that  the  synchronous  motor  can  supply?     What 
is  the  necessary  excitation  voltage  of  the  synchronous  motor  at 
this  time? 

42.  Induction  motors  at   the  end  of  a  3-phase  transmission 
line  deliver  a  total  power  of  3500  h.p.  and  operate  at  a  resultant 
efficiency  and  power  factor  of  91.6  per  cent,  and  0.903  respectively. 
The  line  voltage  is  13,200  volts.     A  1640-kv.-a.   synchronous 
motor  is  operated  in  parallel  with  these  motors  to  improve  the 
power  factor  of  the  load,  and  to  supply  an  additional  load  of 
1000  h.p.     The  armature  windings  of  this  synchronous  motor 
are  connected  in  Y  and  have  an  effective  resistance  of  1.52  ohms 


SYNCHRONOUS  MOTORS  65 

per  phase.     The  rotational  losses  at  normal  voltage  are  57  h.p. 
and  may  be  assumed  to  be  constant. 


Field  current 

Open  circuit  terminal 
voltage 

Short  circuit  armature 
current 

50 
100 
150 
200 

7,500 
10,100 
14,700 
15,800 

75 
155 
227 

250 

16,700 

At  what  per  cent,  of  its  rated  capacity  must  the  synchronous 
motor  be  operated  so  that  the  power  factor  of  the  entire  load  shall 
be  unity  and  the  line  voltage  13,200  volts?  Calculate  the  neces- 
sary field  current  of  the  motor  for  this  condition  by  the  magneto- 
motive-force method. 

43.  A  1000-h.p.  synchronous  motor  is  operating  at  the  end 
of  a  3-phase  transmission  line  which  has  a  resistance  of  0.58  ohm 
and  a  reactance  of  0.64  ohm  per  conductor.     The  motor  is  over- 
excited so  that  it  takes  a  line  current  of  216  amperes  at  a  power 
factor  of  0.82,  and  with  a  line  potential  difference  of  2210  volts. 

What  is  the  line  voltage  at  the  generating  station?  What  is 
the  efficiency  of  transmission? 

44.  A  synchronous  motor  operating  at  the  end  of  a  3-phase" 
transmission  line  takes  a   constant  power  of   1500  kw.     The 
resistance  and  reactance  of  the  line  are  respectively  2.2  ohms  and 
2.6  ohms  per  conductor.     The  line  voltage  at  the  generating 
station  is  maintained  at  5740  volts.     The   excitation  of  the 
motor  is  adjusted  so  that  the  line  loss  has  its  least  possible  value. 

What  is  the  terminal  voltage  at  the  motor?  What  is  the 
efficiency  of  transmission? 

45.  A  synchronous  motor  operating  at  the  end  of  a  3-phase 
transmission  line  which  has  a  resistance  of  3.6  ohm  and  a  re- 
actance of  4.1  ohm  per  conductor  takes  a  constant  power  of 
720  kw.  from  the  line  and  the  excitation  is  adjusted  so  that  the 
line  voltage  is  5000  volts.     The  line  voltage  at  the  generating 
station  is  also  maintained  at  5000  volts. 

(a)  What   current   does   the   motor   take?     At   what  power 
factor  is  it  operating? 

(b)  If  the  motor  is  Y-connected  and  has  an  effective  resist- 
ance of  0.40  ohm  and  a  synchronous  reactance  of  15.4  ohms  per 
phase,  what  is  the  necessary  excitation  voltage? 


66     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

46.  A  1200-h.p.  synchronous  motor  is  operated  at  the  end  of 
a  3-phase  transmission  line  which  has  a  resistance  of  3.3  ohms 
and  a  reactance  of  3.7  ohms  per  conductor.     The  line  voltage 
at  the  generating  station  is  5000  volts,  and  the  motor  delivers 
energy  to  a  constant  load  of  1000  h.p.     The  rotational  losses  of 
the  motor  are  29.8  kw.  and  may  be  assumed  constant.     The 
armature  windings  are  connected  in  Y  and  have  a  resistance  of 
0.40  ohm  and  a  synchronous  reactance  of  15.4  ohms  per  phase. 

If  the  maximum  allowable  line  current  is  135  amperes  wnat  is 
the  greatest  possible  terminal  voltage  at  the  motor?  What  is 
the  necessary  excitation  voltage  of  the  motor? 

47.  At  the  end  of  a  transmission  line  which  has  a  resistance  of 
12.2  ohms  and  a  reactance  of  17.4  ohms  per  conductor  there  is  a 
synchronous  motor  which  delivers  a  constant  load  of  1300  h.p. 
The  armature  windings  of  the  motor,  which  are  connected  in  Y, 
have  an  effective  resistance  of  0.94  ohm  and  a  synchronous 
reactance  of  56  ohms  per  phase.     At  no  load  with  an  impressed 
voltage  of  11,000  volts  the  motor  takes  a  current  of  1.52  amperes 
when  the  field  current  is  adjusted  for  unit  power  factor.     The 
line  voltage  at  the  generating  station  is  maintained  constant  at 
11,500  volts. 

If  the  line  current  is  limited  to  120  per  cent,  of  its  full-load 
value,  which  is  62.5  amperes,  what  is  the  least  additional  react- 
ance that  should  be  inserted  in  each  line  so  that  a  terminal  poten- 
tial difference  of  12,000  volts  may  be  obtained  at  the  motor? 
What  is  the  necessary  excitation  voltage  of  the  motor  in  this 
case? 

48.  A  13,800-volt,  3-phase  synchronous  motor  has  a  full-load 
capacity  of  1350  h.p.     The  armature  windings  are  connected 
in  Y  and  have  an  effective  resistance  of  2.18  ohms  and  a  syn- 
chronous reactance  of  92  ohms  per  phase.     At  no  load  with  an 
impressed  voltage  of  13,800  volts  the  motor  takes  28.2  kw.  at 
unit  power  factor  when  the  excitation  is  normal.     This  motor 
is  operated  at  the  end  of  a  transmission  line  which  has  a  resist- 
ance of  22  ohms  and  a  reactance  of  28  ohms  per  conductor. 
The  line  voltage  at  the  generating  station  is  maintained  con- 
stant at  14,000  volts. 

When  the  motor  is  delivering  1200  h.p.  what  is  the  greatest 
potential  difference  at  the  motor  if  the  armature  current  is 
limited  to  125  per  cent,  of  its  full-load  value?  What  is  the  neces- 
sary excitation  voltage  of  the  motor? 


SYNCHRONOUS  MOTORS  67 

49.  At  the  end  of  a  3-phase  transmission  line  which  has  a 
resistance  of  0.62  ohm  and  a  reactance  of  0.64  ohm  per  conductor 
there  is  a  synchronous  motor  whose  armature  windings  are  con- 
nected in  Y  and  have  an  effective  resistance  of  0.172  ohm  and  a 
synchronous  reactance  of  2.12  ohms  per  phase.     The  rotational 
losses  of  the  motor  are  17.2  kw.  and  may  be  assumed  constant. 

When  instruments  in  the  generating  station  indicate  that  the 
transmission  line  is  receiving  780  kw.  at  2250  volts  and  a  power 
factor  of  0.88  (leading),  what  is  the  output  of  the  motor?  What 
is  the  line  voltage  at  the  motor? 

50.  A  3-phase  transmission  line  has  a  resistance  of  12.6  ohms 
and  a  reactance  of  16.4  ohms  per  conductor.     The  generating 
station  which  delivers  energy  to  this  line  maintains  a  constant 
line  potential  difference  of  14,000  volts.     At  the  end  of  the  line 
there  is  a  synchronous  motor  whose  armature  windings  are 
connected  in  Y  and  have  an  effective  resistance  of  1.52  ohms 
and  a  synchronous  reactance  of  37.4  ohms  per  phase.     The 
rotational  losses  of  this  motor  are  36.8  kw.  and  may  be  assumed 
constant.     When  the  motor  is   delivering  2000   h.p.   and  the 
excitation  voltage  is  adjusted  to  its  greatest  value  of  16,700 
volts  what  is  the  terminal  voltage  of  the  motor?     At  what 
power  factor  is  the  motor  operating? 


CHAPTER  IV 
INDUCTION  MOTORS 

1.  A  2-phase  induction  motor  has  a  stator  with  2  slots  per 
pole  per  phase.     There  are  4  inductors  in  series  per  slot.     The 
coil  pitch  is  4  slots.     The  effective  value  of  the  current  in  each 
phase  of  the  stator  winding  is  7.07  amperes,  and  the  current 
in  phase  two  lags  behind  the  current  in  phase  one  by  90  electrical 
degrees.     Place  the  first  slot  in  phase  one  at  the  extreme  left 
of  the  paper  and  draw  the  zero  lines  to  allow  for  a  maximum 
ordinate  of  2  in.     Use  the  following  scales  in  these  plots. 

Abscissae Pole  pitch  =  3  in. 

Ordinates 40  ampere  turns  =  1  in. 

(a)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  due  to  the  stator  currents  at  the  time  that  the  current  in 
phase  one  is  a  maximum. 

(b)  Plot  the  distribution  of  magnetomotive  force  in  the  air- 
gap  at  one-eighth  of  a  period  later  than  in  (a) . 

(c)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  at  one-quarter  of  a  period  later  than  in  (a). 

2.  A  2-phase  induction  motor  has  a  stator  with  4  slots  per 
pole  per  phase.     There  are  4  inductors  in  series  per  slot.     The 
coil  pitch  is  8  slots.     The  effective  value  of  the  current  in  each 
phase  of  the  stator  winding  is  7.07  amperes,  and  the  current  in 
phase  two  lags  behind  the  current  in  phase  one  by  90  electrical 
degrees.     Place  the  first  slot  in  phase  one  at  the  extreme  left  of 
the  paper  and  draw  the  zero  lines  to  allow  for  a  maximum  or- 
dinate of  3  in.     Use  the  following  scales  in  this  plots : 

Abscissae Pole  pitch   =  3  in. 

Ordinates 40  ampere  turns  =   1  in. 

(a)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  due  to  the  stator  currents  at  the  time  that  the  current  in 
phase  one  is  a  maximum. 

Note.  Unless  otherwise  stated  assume  that  the  ratio  of  the  turns  in  the 
stator  and  rator  windings  is  the  same  as  the  ratio  of  transformation. 

68 


INDUCTION  MOTORS  69 

(b)  Plot  the  distribution  of  magnetomotive  force  in  the  air-gap 
at  one-eighth  of  a  period  later  than  in  (a). 

(c)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  at  one-quarter  of  a  period  later  than  in  (a). 

3.  A  3-phase  induction  motor  has  a  stator  with  2  slots  per 
pole  per  phase.     There  are  4  inductors  in  series  per  slot.     The 
coil  pitch  is  6  slots.     The  effective  value  of  the  current  in  each 
phase  of  the  stator  winding  is  7.07  amperes,  and  the  current 
in  phase  two  lags  behind  the  current  in  phase  one  by  120  elec- 
trical degrees  and  leads  the  current  in  phase  three  by  the  same 
amount.     Place  the  first  slot  in  phase  one  at  the  extreme  left 
of  the  paper  and  draw  the  zero  lines  to  allow  for  a  maximum 
ordinate  of  2  in.     Use  the  following  scales  in  these  plots: 

Abscissae Pole  pitch  =  3  in. 

Ordinates 40  ampere  turns  =  1  in. 

(a)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  due  to  the  stator  currents  at  the  time  that  the  current  in 
phase  one  is  a  maximum. 

(b)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  at  one-eighth  of  a  period  later  than  in  (a). 

(c)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  at  one-quarter  of  a  period  later  than  in  (a). 

4.  A  3-phase  induction  motor  has  a  stator  with  4  slots  per  pole 
per  phase.     There  are  4  inductors  in  series  per  slot.     The  coil 
pitch  is  12  slots.     The  effective  value  of  the  current  in  each 
phase  of  the  stator  winding  is  7.07  amperes,  and  the  current  in 
phase  two  lags  behind  the  current  in  phase  one  by  120  electrical 
degrees  and  leads  the  current  in  phase  three  by  the  same  amount. 
Place  the  first  slot  in  phase  one  at  the  extreme  left  of  the  paper 
and  draw  the  zero  lines  to  allow  for  a  maximum   ordinate  of 
4  in.     Use  the  following  scales  in  these  plots: 

Abscissas Pole  pitch  =   3  in. 

Ordinates 40  ampere  turns  =   1  in. 

(a)  Plot  the  distribution  of  the  magnetomotive  force  in  the  air- 
gap  due  to  the  stator  currents  at  the  time  that  the  current  in  phase 
one  is  a  maximum. 

(b)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  at  one-eighth  of  a  period  later  than  in  (a). 


70     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

(c)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  at  one-quarter  of  a  period  later  than  in  (a). 

6.  A  3-phase  induction  motor  has  a  stator  with  4  slots  per  pole 
per  phase.  There  are  4  inductors  per  slot,  two  of  which  are  in 
one  coil  and  two  in  another.  The  coil  pitch  is  10  slots.  There  are 
two  coil  sides  in  each  slot.  The  effective  value  of  the  current  in 
each  phase  of  the  stator  winding  is  7.07  amperes,  and  the  current 
in  phase  two  lags  behind  the  current  in  phase  one  by  120  electrical 
degrees  and  leads  the  current  in  phase  three  by  the  same  amount. 
Place  the  first  slot  in  phase  one  at  the  extreme  left  of  the  paper  and 
draw  the  zero  lines  to  allow  for  a  maximum  ordinate  of  4  in 
Use  the  following  scales  in  these  plots: 

Abscissae Pole  pitch  =  3  in. 

Ordinates 40  ampere  turns  =  1  in. 

(a)  Plot  the  distribution  of  the  magnetomotive  force  in  the  air- 
gap  due  to  the  stator  currents  at  the  time  that  the  current  in  phase 
one  is  a  maximum. 

(b)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  at  one-eighth  of  a  period  later  than  in  (a). 

(c)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  at  one-quarter  of  a  period  later  than  in  (a). 

6.  Assume  that  the  stator  winding  of  a  3-phase  induction  motor 
is  uniformly  distributed.     The  coil  pitch  is  unity  and  the  phase 
spread  is  one-third  the  polar  pitch.     The  current  in  phase  two 
lags  behind  the  current  in  phase  one  by  120  degrees  and  leads 
the  current  in  phase  three  by  the  same  amount.     Place  the  be- 
ginning of  phase  one  at  the  extreme  left  of  the  paper  and  draw 
the  zero  lines  to  allow  for  a  maximum  ordinate  of  3  in.     As 
ordinates  let  1  in.  equal  the  maximum  ampere  turns  per  pole 
per  phase,  and  as  abscissae  let  3  in.  equal  the  pole  pitch. 

(a)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  at  the  time  that  the  current  in  phase  one  is  a  maximum. 

(b)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  at  one-eight  of  a  period  later  than  in  (a) . 

(c)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  at  a  time  one-quarter  of  a  period  later  than  in  (a). 

7.  Assume  that  the  stator  winding  of  a   3-phase  induction 
motor  is  uniformly  distributed.     The  coil  pitch  is   unity  and 
the  phase  spread  is  two-thirds  the  polar  pitch.     The  current  in 
phase J;wo  lags  behind  the  current  in  phase  one  by  120  degrees 


INDUCTION  MOTORS  71 

and  leads  the  current  in  phase  three  by  the  same  amount.  Place 
the  beginning  of  phase  one  at  the  extreme  left  of  the  paper  and 
draw  the  zero  lines  to  allow  for  a  maximum  ordinate  of  3  in.  As 
ordinates  let  1  in.  equal  the  maximum  ampere  turns  per  pole  per 
phase,  and  as  abscissae  let  3  in.  equal  the  pole  pitch. 

(a)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  at  the  time  that  the  current  in  phase  one  is  a  maximum. 

(b)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  at  one-eighth  of  a  period  later  than  in  (a). 

(c)  Plot  the  distribution  of  the  magnetomotive  force  in  the 
air-gap  at  a  time  one-quarter  of  a  period  later  than  in  (a) . 

8.  At  no  load  a  25  h.p.  induction  motor  takes  865  watts  at 
0.18  power  factor  from  a  250- volt,  3-phase  circuit.     The  effective 
resistance  of  the  stator  winding  is  0.15  ohms  per  phase.     Neg- 
lect the  leakage  reactance.     What  would  have  been  the  no-load 
power  and  power"  factor  if  the  motor  had  been  designed  with 
an  air-gap  which  would  have  made  the  reluctance  of  the  magnetic 
circuit  25  per  cent,  greater? 

9.  A  200-h.p.  induction  motor  is  designed  to  operate  from  a 
3-phase,  980-volt,   20-cycle  circuit.     Under  this  condition  the 
no-load  power  and  power  factor  are  6.62  kw.  and  0.086.     The 
friction  and  windage  loss  is  4.0  kw.     The  stator  winding  is  delta- 
connected  and  has  an  effective  resistance  of  0.387  ohms  per 
phase.     What  will  be  the  no-load  power  and  power  factor   if 
this  motor  is  operated  from  a  3-phase,  1100-volt,  25-cycle  cir- 
cuit?    Neglect   the   leakage   reactance   and   assume   that   the 
core  losses  vary  as  B1'7/1'3 

10.  At  no  load  a  570-h.p.,  3-phase   induction   motor   takes 
15.1   kw.   at  0.081   power  factor.     The   friction   and  windage 
loss  is  12.0  kw.     The  effective  resistance  of  the  stator  winding 
is  0.757  ohm  per  phase.     The  windings  are  connected  in  delta. 
There  are  8  inductors  in  series  per  slot.     If  the  motor  is  rewound 
with  the   same  size   wire   but  with  6   inductors  in  series  per 
slot,  what  no-load  current  and  power  will  it  take  from  the  same 
circuit?     Neglect  the  leakage  reactance  and  assume  that  the  core 
losses  vary  as  B1'7  /1-3. 

11.  Both  the  stators  and  rotors  of  two   150-h.p.,  3-phase  in- 
duction motors  are  of  the  same  design  except  that  one  is  wound 
for  6  poles  and  the  other  for  8  poles.     Each  stator  has  the  same 
number  of  slots  with  the  same  number  of  inductors  per  slot. 
The  six-pole  motor  takes  5.7  kw.  from  a  500  volt,  38-cycle  circuit 


72     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

at  no  load  and  the  line  current  is  46  amperes.  The  friction  and 
windage  loss  is  3.9  kw.  The  effective  resistance  of  the  stator 
winding  is  0.082  ohm  per  phase.  The  effective  resistance  of  the 
stator  winding  of  the  8-pole  motor  is  11  per  cent,  less  than  this. 
The  windings  are  connected  in  Y.  What  current  and  power 
will  the  8-pole  motor  take  at  no  load  from  the  same  circuit? 
Assume  that  the  friction  and  windage  loss  varies  as  the  speed 
and  neglect  the  leakage  reactance.  Where  approximations  are 
made  state  what  they  are. 

12.  At  no  load  a  335-h.p.,  3-phase  induction  motor  takes  a 
line  current  of  15.5  amperes  and  absorbs  10.1  kw.  from  a  2000- 
volt,  50-cycle  circuit.     The  effective  resistance  and  the  leakage 
reactance  of  the  stator  winding  are  respectively,  0.248  and  0.76 
ohm  per  phase.     The  windings  are  connected  in  Y.     The  friction 
and  windage  loss  is  2.5  kw.     What  current  and  power  will  this 
motor  take  at  no  load  from  a  2200-volt,  60-cycle  circuit?     As- 
sume that  the  core  losses  vary  as  B1'1} 1-3  and  that  the  generated 
voltage  due  to  the  air-gap  flux  is  equal  to  the  impressed  voltage 
less  the  magnetizing  current  multiplied  by  the  leakage  reactance. 

13.  At  no  load  a  150-h.p.,  3-phase  induction  motor  takes  a 
line  current  of  46  amperes  and  absorbs  5.65  kw.  from  a  500- volt, 
38-cycle  circuit.     The  friction  and  windage  loss  is  4.1  kw.     The 
stator  has  12  slots  per  pole  and  the  winding  pitch  is  12  slots. 
If  this  motor  is  rewound  with  the  same  number  of  turns  using 
a  pitch  of  10  slots,  what  current  and  power  will  it  take  at  no 
load  from  the  same  circuit?     Neglect  the   resistance  and  the 
leakage  reactance,  and  assume  that  the  core  losses  vary  as  B1' 7  / 1-  3 

14.  The   stator  of  a   150-h.p.,   3-phase,   500-volt  induction 
motor  has  108  slots,  and  in  each  of  the  following  cases  it  is  wound 
for  6   poles  with  4  inductors  per  slot.     With  a  winding  pitch 
of  18  slots  the  core  losses  are  3.9  kw.  and  the  magnetizing  cur- 
rent is  45  amperes.     If  a  winding  pitch  of  14  slots  is  used  what 
will  be  the  core  losses  and  the  magnetizing  current  at  the  same 
voltage  and  frequency? 

15.  The    stator    of   a    25-h.p.,    3-phase,    250-volt   induction 
motor  has  72  slots,  and  in  each  of  the  following  cases  it  is  wound 
with  8  inductors  per  slot.     If  this  motor  is  wound  for  8  poles 
with  a  winding  pitch  of  8  slots  the  core  losses  are  590  watts, 
the  friction  and  windage  loss  is  220  watts,  and  the  magnetizing 
current  is  11  amperes.     What  current  and  power  will  this  motor 
take  from  the  same  circuit  at  no  load  if  it  is  wound  for  6  poles 


INDUCTION  MOTORS  73 

with  a  winding  pitch  of  10  slots?  Neglect  the  resistance  and 
leakage  reactance  and  assume  that  the  friction  and  windage  loss 
varies  as  the  speed  and  that  the  core  losses  vary  as  B  i-7/1-3. 

16.  The  stator  of  a  25-h.p.,  3-phase,  50-cycle  induction  motor 
is  wound  for  250  volts.     With  an  impressed  voltage  of  110  volts 
the  starting  current  is  61  amperes  at  0.336  power  factor.     The  ef- 
fective resistance  and  leakage  reactance  of  the  stator  winding  are 
respectively    0.159  ohm  and  0.46  ohm  per  phase.     The  stator 
has  72  slots  with  9  inductors  per  slot  and  the  rotor  has  120  slots 
with  2  inducters  per  slot.     Both  the  stator  and  rotor  windings 
connected  in  Y.     The  ohmic  resistance  of  the  rotor  winding  is 
0.015  ohm  per  phase. 

What  is  the  leakage  inductance  of  the  rotor  winding  per  phase? 
What  is  the  ratio  of  effective  to  ohmic  resistance  for  the  rotor 
winding  at  50  cycles? 

If  the  stator  is  rewound  for  500  volts  by  using  twice  as 
many  turns  of  wire  of  one-half  the  size,  what  voltage  should 
be  impressed  to  have  a  starting  current  of  30  amperes? 

17.  The  full-load  line  current  taken  by  a  200-h.p.,  3-phase,  980- 
volt  induction  motor  is  101  amperes.     The  stator  has  216  slots 
with  5  inductors  per  slot,  and  the  ohmic  resistance  of  the  winding 
is  0.246  ohm  per  phase.     The  rotor  has  288  slots  with  1  inductor 
per  slot  and  the  ohmic  resistance  of  the  winding  is  0.010  ohm  per 
phase.     Both  the  stator  and  rotor  windings  are  connected  in  delta. 
The  leakage  inductances  of  the  stator  and  rotor  windings  are 
respectively  3.9  mil-henrys  and  0.19  mil-henry  per  phase.     At 
the  rated  frequency  of  50  cycles  the  ratios  of  effective  to  ohmic 
resistance  are  respectively  1.6  and  1.9  for  the  stator  and  rotor. 

(a)  What  voltage  should  be  impressed  on  this  motor  in  order 
that  the  starting  current  will  be  twice  the  full-load  current? 

(b)  If  the  stator  winding  is  reconnected  in  Y  what  voltage 
should  be  impressed  on  the  motor  in  order  that  the  starting  cur- 
rent will  be  twice  the  full-load  current ?    In  this  case  the  rated  volt- 
age becomes  1700  volts. 

18.  At  full  load  the  stator  and  rotor  copper  losses  of  a  500-h.p., 
3-phase,  2000-volt  induction  motor  are   respectively   4.14  per 
cent,  and  1.82  per  cent.     The  core  loss  is  2470  watts  and  the  fric- 
tion and  windage  loss  is  11.0  kw.     The  magnetizing  component 
of  the  line  current  is  23  amperes.     At  the  rated  frequency  of  60 
cycles  the  ratio  of  effective  resistance  to  ohmic  resistance  is  1.55 
for  the  stator  winding  and  1.85  for  the  rotor  winding.     At  the 


74     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

rated  frequency  the  leakage  reactances  of  the  windings  are  3.2 
times  their  effective  resistances.  What  line  voltage  should  be 
impressed  on  this  motor  when  starting  to  give  a  line  current  of  200 
amperes? 

19.  The  ohmic  resistances  of  the  stator  and  rotor  windings  of 
an  indication  motor  are  respectively  0.04  ohm  and  0.01  ohm. 
At  normal  freqency  the  ratios  of  effective  resistance  to  ohmic 
resistance  are  respectively  1.5  and  1.8  for  the  stator  and  rotor 
windings  and  the  leakage  reactances  are  3.1  times  their  effective 
resistances.     The  stator  winding  has  108  inductors  per  phase  and 
the  rotor  winding  has  54  inductors  per  phase.     Assume  that  the 
core  loss  due  to  leakage  flux  and  the  increase  in  the  resistance 
due  to  a  non-uniform  distribution  of  current  over  the  cross- 
section  of  the  inductors  both  vary  directly  as  the  frequency. 

(a)  If  the  voltage  impressed  on  an  induction  motor  when 
starting  is  increased  30  per  cent.,  how  much  is  the  starting  current 
increased?     How  much  is  the  starting  torque  increased? 

(b)  If  the  frequency  of  the  voltage  impressed  on  an  induction 
motor  when  starting  is  increased  10  per  cent.,  how  much  is  the 
starting  current  decreased?     How  much  is  the  starting  torque 
decreased. 

20.  A  570-h.p.  induction  motor  is  designed  to  receive  power 
from  a  3-phase,   1900-volt,  22.5-cycle  circuit.     The  ohmic  re- 
sistances of  the  stator  and  rotor  windings  are  0.488  ohm  and 
0.0138  ohm  per  phase,  and  at  the  rated  frequency  the  effective 
resistances  are  1.3  and  1.5  times  as  great.     The  leakage  reactances 
at  the  rated  frequency  are  three  times  the  effective  resistances. 
The  stator  has  864  inductors  per  phase  and  the  rotor  144  in- 
ductors per  phase.     The  motor  is  wound  for  36  poles,  and  both  of 
the  windings  are  connected  in  delta.     The  friction  and  windage 
loss  is  12  kw. 

What  is  the  slip  at  full  load  if  the  generated  voltage  in  the  stator 
winding  due  to  the  air-gap  flux  is  93  per  cent,  of  the  impressed 
voltage?  What  is  the  starting  torque  with  full  voltage  impressed 
on  the  stator  windings? 

21.  The  effective  resistances  of  the  stator  and  rotor  windings  of 
a  3-phase  inducton  motor  are  0.071  ohm  and  0.0218  ohm  per 
phase,  and  the  leakage  reactances  at  normal  frequency  are  0.22 
ohm  and  0.066  ohm  per  phase.     The  stator  has  108  slots  with  5 
inductors  per  slot,  and  the  rotor  has  126  slots  with  2  inductors 
per  slot.     Both  of  the  windings  are  connected  in  Y.     At  starting, 


INDUCTION  MOTORS  75 

with  full  impressed  voltage,  the  current  is  3.6  times  the  full-load 
value  and  the  torque  is  0.62  of  the  full-load  torque.  What  will 
be  the  starting  current  and  the  starting  torque  with  full  impressed 
voltage  if  resistances  of  0.2  ohm  are  inserted  in  each  phase  of  the 
rotor  winding? 

22.  With  full-load  current  the  slip  and  brake  torque  of  a  500- 
h.p.,  3-phase,  60-cycle  induction  motor  are  1.82  per  cent,  and 
16,300  pound-feet.     The  ohmic  resistance  and  leakage  inductance 
of  the  rotor  winding  are  0.24  ohm  and  3.1  mil-henrys  per  phase. 
What  will  be  the  slip,  the  brake  torque  and  the  output  when  the 
current  has  its  full-load  value  if  resistances  of  2  ohms  are  inserted 
in  each  phase  of  the  rotor  winding?     Estimate  the  copper  loss 
in  each  of  these  additional  resistance  units. 

23.  The  resistance  and  leakage  inductance  of  the  rotor  wind- 
ings of  a  3-phase,  38-cycle  induction  motor  are  0.013  ohm  and 
0.27    mil-henry  per  phase.     The  full-load  torque  and  slip  are 
2,190  Ib.-ft.  and  2.6  per  cent. 

(a)  To  what  per  cent,  of  its  normal  full-load  value  should  the 
air-gap  flux  be  reduced  so  that  the  motor  will  deliver  its  full- 
load  torque  at  one-half  the  full-load  speed? 

(b)  What  resistance  should  be  inserted  in  each  phase  of  the 
rotor  so  that  the  motor  will  deliver  its  full-load  torque  at  one- 
half  the    full-load  speed?     Assume  that  the  air-gap  flux  has 
its  normal  value. 

(c)  Compare  the  rotor  currents  in  (a)  and  (b). 

24.  The  full-load  distribution  of  losses  in  a  10-h.p.,  220- volt, 
3-phase  induction  motor  is: 

Stator  copper  loss 3.9  per  cent. 

Rotor  copper  loss 4.8  per  cent. 

Core  loss 3.1  per  cent. 

Friction  and  windage  loss 4.0  per  cent. 

(a)  What  are  the  slip  and  efficiency  at  full  load? 

(b)  What  are  the  slip  and  efficiency  at  one-half  of  full  load? 

25.  At  full-load  the  slip  of  a  335-h.p.,  2000-volt,  50-cycle,  6-pole 
induction  motor  is  1.8  per  cent.     The  ohmic  resistances  of  the 
stator  and  rotor  windings  are  0.165  ohm  and  0.0127  ohm  per 
phase,  and  the  leakage  inductances  are  2.4  mil-henrys  and  0.22 
mil-henry  per  phase.     Both  of  the  windings  are  connected  in  Y. 
The  stator  has  72  slots  with  10  inductors  per  slot,  and  the  rotor 
has  90  slots  with  2  inductors  per  slot.     The  ratios  of  effective 


76     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

resistance  to  ohmic  resistance  at  50  cycles  are  1.45  and  1.75  for 
the  stator  and  rotor  windings.  At  full  load  the  voltage  gener- 
ated in  the  stator  winding  by  the  mutual  flux  is  93  per  cent,  of 
the  impressed  voltage. 

To  what  per  cent,  of  its  full-load  value  should  the  air-gap 
flux  be  reduced  so  that  the  starting  current  will  be  175  amperes? 
What  is  the  starting  torque  for  this  condition?  What  are  the 
full-load  torque  and  rotor  current? 

26.  In  problem  25  what  resistance  should  be  inserted  in  each 
phase  of  the  rotor  winding  so  that  when  starting  with  full  im- 
pressed voltage  the  rotor   current   will  be  twice  its  full-load 
value?     What  is  the  starting  torque  for  this  condition?     What 
is  the  running  torque  with  full-load  current? 

27.  In  problem  25  what  reactance  should  be  inserted  in  each 
phase  of  the  rotor  winding  so  that  when  starting  with  full  im- 
pressed voltage  the  rotor   current    will  be  twice  its  full-load 
value?     What  is  the  starting  torque  for  this  condition?     What 
is  the  running  torque  with  full-load  current? 

28.  At  the  instant  of  starting  on  a  reduced  voltage  of  500  volts 
a  500-h.p.,  3-phase,  60-cycle  induction  motor  takes  a  line  current 
of   152  amperes  at  0.31  power  factor.     The  starting  torque  is 
790    pound-feet.      The    motor    is    wound    for    44   poles.      The 
ohmic  resistance  of  the  rotor  winding  is  0.0306  ohm  measured 
between   terminals — when    the  winding  is  not  short-circuited. 
The  effective  resistance  is  1.7  times  as  great  at  the  rated  fre- 
quency.    The    stator    winding    has    704    inductors    per   phase 
and  the  rotor  winding  has  220  inductors  per  phase.     The  fric- 
tion and  windage  loss  is  11.  kw.     With  an  impressed  voltage 
of  2000  volts  what  brake  torque  would  be  delivered  when  the 
slip  is  1.8  per  cent?     Assume  that  the  voltage  generated  in  the 
stator  winding  by  the  air-gap  flux  is  94  per  cent,  of  the  im- 
pressed voltage. 

29.  A  570-h.p.,  3-phase  induction  motor  is  arranged  so  that 
it  may  be  connected  to  the  line  through  a  compensator  at  start- 
ing.    The  motor  is  wound  for  36  poles  and  a  line  voltage  of  1900 
volts  at  22.5  cycles.     When  the  compensator  reduces  the  im- 
pressed to  600  volts  the  line  current  is  200  amperes  and  the  power 
is  58.8  kw.  at  the  instant  of  starting.     The  stator  and  rotor  ohmic 
resistances  are  equal  when  reduced  to  the  same  side  and  the  ef- 
fective resistances  are  1.5  times  as  great  as  the  ohmic.     The 
friction  and  windage  loss  at  full  load  is  12.  kw.     Assume  that 


INDUCTION  MOTORS  77 

the  voltage  generated  in  the  stator  winding  at  full  load  is  94 
per  cent,  of  the  impressed  voltage.  What  is  the  slip  when  the 
motor  delivers  its  rated  load?  What  is  the  starting  torque  with 
600  volts  impressed  on  the  motor? 

30.  At  full  load  the  slip  of  a  500-h.p.,  3-phase  induction  motor 
is  1.80  per  cent.     The  motor  is  wound  for  12  poles  and  an  im- 
pressed voltage  of  2200  volts  at  25  cycles.     When  the  rotor  wind- 
ing is  not  short  circuited  the  hot  resistance — referred  to  the  stator 
—measured  between  terminals  is  0.431  ohm,  and  the  effective 
resistance  is  1.65  times  as  much.     What  is  the  starting  torque 
when  the  impressed  voltage  is  adjusted  so  that  the  rotor  cur- 
rent is  twice  its  full-load  value? 

31.  At  full  load  the  slip  of  a  1000-h.p.,  3-phase  induction  motor 
is  1.72  per  cent.     The  motor  is  wound  for  12  poles  and  an  im- 
pressed voltage  of  2200  volts  at  25  cycles.     When  the  rotor  wind- 
ing is  not  short  circuited  the  hot  resistance  measured  between 
terminals  is  0.0895  ohm,  and  the  effective  resistance  is  1.6  times 
as  great.     The  ratio  of  transformation  from  stator  to  rotor  is 
2200  to  1500.     The  effective  resistance  and  leakage  reactance  of 
the  stator  winding  between  terminals  are  0.195  ohm  and  0.59 
ohm. 

What  voltage  should  be  impressed  on  this  motor  so  that  the 
starting  torque  will  be  the  same  as  the  full-load  torque?  What 
is  the  starting  current  for  this  condition  and  how  does  it  compare 
with  the  full-load  current? 

32.  In  problem  31  what  resistance  should  be  inserted  in  each 
phase  of  the  rotor   winding   in  order  that  the  starting  torque 
for   the   rated   voltage   will  be  equal  to  the  full-load  torque? 
What  is  the  starting  current  for  this  condition  and  how  does 
it   compare  with  the  full-load  current? 

33.  The  ohmic  resistances  of  the  stator  and  rotor  windings  of  a 
500-h.p.,  3-phase  induction  motor  are  0.24  ohm  and  0.0153  ohm 
per  phase,  and  the  effective  resistances  at  the  rated  frequency  are 
respectively  1.5  and  1.6  times  as  great.     The  leakage  reactances 
at  the  rated  frequency  are  respectively  1.1  ohms  and  0.075  ohm 
per  phase.     The  motor  is  wound  for  44  poles  and  an  impressed 
voltage  of  2000  volts  at  60  cycles.     The  stator  has  2112  inductors 
and  the  rator  660  inductors.     Both  windings  are  connected  in  Y. 
At  full  load  the  voltage  generated  in  the  stator  winding  by  the 
air-gap  flux  is  94  per  cent,  of  the  impressed  voltage. 

What  resistance  should  be  inserted  in  each  phase  of  the  rotor 


78     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

in  order  that  the  starting  torque  with  full  impressed  voltage  may 
have  its  maximum  value?  What  is  this  torque?  What  is  the 
starting  current  for  this  condition  and  how  does  it  compare  with 
the  full-load  current? 

34.  The  ohmic  resistances  of  the  stator  and  rotor  windings 
of  a  150-h.p.,  3-phase  railway  induction  motor  are  0.054  ohm  and 
0.014  ohm  per  phase,  and  the  effective  resistances  at  the  rated 
frequency  are  respectively  1.45  and  1.65  times  as  great.  The 
leakage  reactances  at  the  rated  frequency  are  respectively  0.24 
ohm  and  0.071  ohm  per  phase.  The  motor  is  wound  for  6  poles 
and  an  impressed  voltage  of  500  volts  at  38  cycles.  The  ratio  of 
transformation  from  stator  to  rotor  winding  is  15  to  7,  and 
both  windings  are  connected  in  Y.  With  full  impressed  voltage 
the  measured  slip  for  a  brake  torque  of  1470  pound-feet  is  2.26 
per  cent. 

What  resistance  should  be  inserted  in  each  phase  of  the  rotor 
winding  so  that  the  starting  torque  with  full  impressed  voltage 
will  have  its  maximum  value?  What  is  this  torque? 


INDUCTION  MOTOR  DATA 


A 

B 

C 

D 

E 

Horse-powc 
Line  voltag 
Type  of  wii 
Frequency  . 
Poles  

>r  

25 
250 
Y 
50 
8 
0.096 

120 
500 
Y 
38 
6 
0.04 
0.01 
2  :  1 

200 
980 
A 
20 
24 
0.246 

335 
2000 
Y 
50 
6 
0.165 
0.0127 
4  :  1 

570 

1900 
A 
22.5 
36 
0.488 
0.0138 
6  :  1 

e 

iding  

Ohmic  resig 
per  phase. 
Ratio  of  in 
Magnetizin 
Core  loss 

tance    (  Stator.  . 
\  Rotor.  . 
insformation 

g  current  (line).. 

11 
590 
220 

39 
1120 
4000 

Friction  an 
No  load 

Blocked..  - 

d  windage  loss  .  .  . 
Line  voltage  .  .  . 
Line  current.  .  . 
Power  

500 
34.5 

4400 
500 
800 
220,000 

2000 
15.3 
10,100 
440 
170 
40,500 

1900 
57.6 
17,200 
600 
200 
65,000 



220 
173 

20,400 

Line  voltage.  .  . 
Line  current  .  .  . 
Power  

100 
57 
3,100 

35.  Draw  the  Heyland  diagram  for  motor  No.  A.     (a)  What 
are  the  full-load  power  factor,  slip  and  efficiency,     (b)  What  is 


INDUCTION  MOTORS  79 

the  break-down  torque?     (c)  At  what  load  is  the  power  factor  a 
maximum? 

36.  Draw  the  Heyland  diagram  for  motor  No.  B.     (a)  What 
are  the  power  factor,  the  slip,  and  the  efficiency  at  full  load? 
(b)  at  one-half  of  full  load?     (c)  What  is  the  break-down  torque? 
(d)  At  what  load  is  the  power  factor  a  maximum? 

37.  Draw  the  Heyland  diagram  for  motor  No.  C.     (a)  What 
are  the  power  factor,  the  slip  and  the  efficiency  at  full  load? 
(b)  What  is  the  break-down  torque?     fc)  At  what  load  is  the 
power  factor  a  maximum? 

38.  Draw  the  Heyland  diagram  for  motor  No.  D.     (a)  What 
are  the  power  factor,  the  slip  and  the  efficiency  at  full  load?     (b) 
at  one-half  full  load?     (c)  What  is  the  break-down  torque? 

39.  Draw  the  Heyland  diagram  for  motor  No.  E.     (a)  What 
are  the  power  factor,  the  slip  and  the  efficiency  at  full  load? 
(b)  What  is  the  break-down  torque?     (c)  At  what  load  is  the 
power  factor  a  maximum? 

40. l  The  following  data  are  given  on  a  5-h.p.,  110- volt,  8- 
pole,  60-cycle,  3-phase  induction  motor.  The  stator  and  rotor 
windings  are  both  connected  in  delta,  and  the  former  has  an  ef- 
fective resistance  of  0.24  ohm  and  a  leakage  reactance  of  0.70  ohm 
per  phase.  The  ohmic  resistance  of  the  rotor  winding  referred 
to  the  stator  is  0.36  ohm  per  phase.  At  no  load  the  motor  takes 
300  watts  at  110  volts  and  a  power  factor  of  0.15. 

When  this  motor  takes  5100  watts  at  110  volts  and  a  power 
factor  of  0.835  what  are  the  slip,  the  power  output  and  the 
torque? 

41. l  A  3-phase,  1000-h.p.  induction  motor  is  taking  916  kw. 
from  a  2200-volt  circuit  at  a  power  factor  of  0.914.  At  no  load 
with  an  impressed  voltage  of  2200  volts  the  line  current  is  75  .1 
amperes  and  the  power  is  15.2  kw.  The  effective  resistance  and 
the  leakage  reactance  of  the  stator  winding,  which  is  Y-connected, 
are  0.118  ohm  and  0.32  ohm  per  phase.  The  ohmic  resistance  of 
the  rotor  winding  referred  to  the  stator  is  0.10  ohm  per  phase. 
What  is  the  output  for  the  specified  load?  What  is  the  slip? 

42. l  A  3-phase,  2600-h.p.  induction  motor  is  taking  2272  kw. 
from  a  6470-volt  circuit,  and  the  line  current  is  245  amperes.  At 
no  load  with  an  impressed  voltage  of  6400  volts  the  line  current 
is  89.5  amperes  and  the  power  is  47.5  kw.  The  ohmic  resistance  of 

1  Use  either  the  transformer  diagram  or  the  equivalent  circuit  in  the 
solution  of  this  problem. 


80     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

the  stator  winding  between  terminals  is  0.561  ohm  at  25°  C.  The 
ohmic  resistance  of  the  open-circuited  rotor  winding  between 
terminals  is  0.0766  ohm  at  25°  C.  The  leakage  reactance  of  the 
stator  winding  is  3.64  ohms  between  terminals.  With  the  rotor 
blocked  375  kw.  is  supplied  when  the  line  current  is  369  amperes. 
The  temperature  of  the  windings  at  this  time  is  25°  C.  Assume 
that  the  ratios  of  effective  resistance  to  ohmic  resistance  are  the 
same  for  the  stator  and  rotor  windings  at  25°  C. 

What  is  the  output  for  the  specified  load?  What  is  the  slip? 
What  is  the  distribution  of  the  losses?  The  temperature  at  this 
time  is  73°  C. 

43. l  A  3-phase,  2000-volt,  60-cycle  induction  motor  has  a  full- 
load  capacity  of  500  h.p.  The  stator  winding  has  an  effective 
resistance  of  0.36  ohm  and  a  leakage  reactance  of  1.1  ohms  per 
phase.  The  rotor  winding  has  an  ohmic  resistance  of  0.157  ohm 
and  a  leakage  inductance  of  1.96  mil-henry s  per  phase  referred 
to  the  stator.  Both  of  the  windings  are  connected  in  Y.  The 
magnetizing  current  is  23  amperes,  the  core  loss  is  2470  watts, 
and  the  friction  and  windage  is  11  kw. 

What  are  the  slip,  the  power  factor,  and  the  efficiency  when  the 
motor  delivers  its  rated  output?  The  impressed  voltage  has 
such  a  value  that  the  voltage  generated  in  the  stator  winding  by 
the  air-gap  flux  is  1850  volts. 

44. l  A  3-phase,  1900-volt,  22.5  cycle,  36-pole  induction  motor 
has  a  full-load  capacity  of  570  h.p.  The  ohmic  resistances  of 
the  stator  and  rotor  windings  are  0.488  ohm  and  0.0138  ohm  per 
phase,  and  their  effective  resistances  at  the  rated  frequency  are 
respectively  1.4  and  1.6  times  as  great.  Both  of  the  windings 
are  connected  in  delta,  and  the  ratio  of  transformation  of  stator 
to  rotor  is  6  to  1.  The  magnetizing  component  of  the  line  cur- 
rent is  57.2  amperes,  the  core  loss  is  2.8  kw.,  and  the  friction 
and  windage  loss  is  12  kw.  With  the  rotor  blocked  the  line 
current  is  200  amperes  when  the  impressed  voltage  is  600  volts. 
Assume  that  the  ratios  of  leakage  reactance  to  effective  resistance 
at  the  rated  frequency  are  the  same  for  both  stator  and  rotor 
windings. 

What  are  the  slip,  the  power  factor,  and  the  efficiency  when  the 
motor  delivers  its  rated  output?  Assume  that  the  voltage 
generated  in  the  stator  winding  by  the  air-gap  flux  is  1790  volts. 

^se  either  the  transformer  diagram  or  the  equivalent  circuit  in  the 
solution  of  this  problem. 


INDUCTION  MOTORS  81 

45. !  A  3-phase,  500-volt,  38-cycle,  6-pole  railway  induction 
motor  has  a  full-load  capactiy  of  120  h.p.  The  ohmic  resistances 
of  the  stator  and  rotor  windings  are  0.04  and  0.01  ohm  per  phase, 
and  their  effective  resistances  at  the  rated  frequency  are  respec- 
tively 1.45  and  1.60  times  as  great.  Both  of  the  windings  are 
connected  in  Y,  and  their  ratio  of  transformation  is  2  to  1.  The 
magnetizing  component  of  the  line  current  is  34  amperes,  and 
the  total  no-load  losses  are  4400  watts.  With  the  rotor  blocked 
the  line  current  is  800  amperes  when  the  impressed  voltage  is  500 
volts.  Assume  that  the  leakage  reactances  of  the  stator  and  rotor 
windings  at  the  rated  frequency  are  equal  when  reduced  to  the 
same  side. 

What  are  the  slip,  the  power  factor  and  the  efficiency  when 
the  motor  delivers  a  torque  of  1200  pound-feet?  Assume  that 
the  voltage  generated  in  the  stator  winding  by  the  air-gap  flux  is 
475  volts. 

46. 1  The  full-load  capacity  of  a  3-phase,  2000-volt,  50-cycle 
induction  motor  is  335  h.p.  The  motor  is  wound  for  6  poles, 
and  both  the  stator  and  rotor  windings  are  connected  in  Y. 
The  effective  resistance  of  the  stator  winding  is  0.23  ohm  and  the 
ohmic  resistance  of  the  rotor  winding  referred  to  the  stator  is 
0.203  ohm  per  phase.  At  50  cycles  the  leakage  reactances  of  the 
windings,  referred  to  the  stator,  are  respectively  0.70  ohm  and 
0.83  ohm  per  phase.  At  no  load  the  motor  takes  15.2  amperes 
from  a  2000-volt  circuit  at  a  power  factor  of  0.187.  The  core  loss 
is  7400  watts  at  this  time.  The  motor  is  operating  with  a  slip  of 
1.43  per  cent.,  and  the  terminal  voltage  has  such  a  value  that 
the  voltage  generated  in  the  stator  winding  is  the  same  as  at 
no  load.  What  is  the  output  of  the  motor?  What  is  the  power 
factor?  What  is  the  efficiency? 

47. l  A  3-phase,  500-volt,  38-cycle  railway  induction  motor  has 
a  full-load  capacity  of  150  h.p.  The  ohmic  resistances  of  the 
stator  and  rotor  windings  are  0.052  ohm  and  0.013  ohm  per  phase, 
and  the  effective  resistances  are  respectively  1.40  and  1.55  times 
as  great  at  the  rated  frequency.  The  motor  is  wound  for  8  poles. 
Both  stator  and  rotor  windings  are  connected  in  Y  and  have  a 
ratio  of  transformation  of  9  to  5.  At  no  load  with  an  impressed 
voltage  of  500  volts  the  motor  takes  a  current  of  46.6  amperes  at 
a  power  factor  of  0.141.  With  the  rotor  blocked  the  motor  takes 

1  Use  either  the  transformer  diagram  or  the  equivalent  circuit  in  the 
solution  of  this  problem. 


82     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

a  current  of  610  amperes  when  the  impressed  voltage  is  500  volts. 
Assume  that  the  leakage  reactances  are  in  the  same  ratio  as  the 
effective  resistances  at  the  rated  frequency. 

What  are  the  slip,  the  power  factor  and  the  efficiency  when  the 
motor  delivers  a  torque  of  1950  pound-feet?  Assume  that  the 
voltage  generated  in  the  stator  winding  by  the  air-gap  flux  is  the 
same  as  at  no  load. 

48. 1  At  no  load  when  the  line  voltage  has  its  rated  value  of  1900 
volts,  a  3-phase,  570-h.p.  induction  motor  takes  a  line  current  of 
57.2  ampers  at  a  power  factor  of  0.091.  The  speed  of  the  rotor  is 
74.6  rev.  per  min.,  and  the  slip  is  one  revolution  in  23  minutes. 
The  ohmic  resistances  of  the  stator  and  rotor  winding  are  0.488 
ohm  and  0.0138  ohm  per  phase.  The  motor  is  wound  for  36 
poles  and  both  the  stator  and  the  rotor  windings  are  connected  in 
delta,  and  have  a  ratio  of  transformation  of  6  to  1.  With  the 
rotor  blocked  the  line  current  is  200  amperes  and  the  power  sup- 
plied is  59.2  kw.  when  the  impressed  voltage  is  600  volts.  Assume 
that  the  ratios  of  effective  to  ohmic  resistance  are  the  same 
for  the  stator  and  rotor,  and  that  the  ratios  of  leakage  reactance 
to  effective  resistance  are  also  the  same  for  each  winding. 

What  is  the  friction  and  windage  loss?     What  is  the  core  loss? 

49. l  A  3-phase,  2000- volt,  60-cycle  induction  motor  is  rated 
to  deliver  500  h.p.  at  full  load.  The  ohmic  resistances  of  the 
stator  and  rotor  windings  are  respectively  0.24  ohm  and  0.0153 
ohm  per  phase,  and  the  ratios  of  effective  to  ohmic  resistance  are 
respectively  1.5  and  1.6.  With  the  rotor  blocked  the  power 
factor  is  0.29.  Assume  that  the  ratios  of  leakage  reactance  to 
effective  resistance  at  the  rated  frequency  are  the  same  for  the 
stator  and  rotor.  The  motor  is  wound  for  44  poles,  and  both 
the  stator  and  rotor  windings  are  connected  in  Y  and  have  a 
ratio  of  transformation  of  16  to  5.  The  friction  and  windage  loss 
is  11.0  kw.  and  the  core  loss  is  2470  watts  at  the  rated  voltage. 

With  the  impressed  voltage  adjusted  so  that  the  voltage  gen- 
erated in  the  stator  winding  by  the  air  gap  flux  is  1120  volts  per 
phase,  what  is  the  slip  in  revolutions  per  minute  at  no  load? 
What  are  the  impressed  voltage,  the  current  and  the  power  for 
this  condition? 

50. 1  The  full-load  capacity  of  a  3-phase,  980-volt,  20-cycle 
induction  motor  is  200  h.p.  The  effective  resistance  and  leakage 

^se  either  the  transformer  diagram  or  the  equivalent  circuit  in  the 
solution  of  this  problem. 


INDUCTION  MOTORS 


83 


reactance  of  the  stator  winding  are  0.342  ohm  and  1.08  ohms  re- 
spectively. The  ohmic  resistance  and  leakage  inductance  of  the 
rotor  winding  referred  to  the  stator  are  0.141  ohm  and  5.1  mil- 
henrys.  The  motor  is  wound  for  24  poles  and  both  the  stator  and 
rotor  windings  are  connected  in  delta.  At  no  load  the  motor 
takes  a  line  current  of  39.4  amperes  and  absorbs  6680  watts  when 
the  impressed  voltage  has  its  rated  value.  The  friction  and  wind- 
age loss  is  4.0  kw. 

If  the  air-gap  flux  is  assumed  to  be  constant  the  maximum 
torque  occurs  when  the  slip  equals  the  ratio  of  the  resistance  of  the 
rotor  winding  to  its  leakage  reactance  at  the  rated  frequency. 
In  this  case  how  much  must  the  impressed  voltage  be  increased 
in  order  that  the  air-gap  flux  will  have  its  no-load  value  when  the 
torque  is  a  maximum? 


F 

G 

H 

' 

J 

Horse-power  

50 

150 

500 

1000 

2600 

Line  voltage  

2200 

500 

2200 

2200 

6400 

Frequency  

60 

60 

25 

25 

25 

Y 

A 

Poles 

12 

12 

36 

Ohmic  resistance  between    /  Stator.  .  . 

6.307 

0.050 

0.357 

0.130 

0.561 

I  Rotor 

0  0286 

0  0772 

0  0766 

hot 

hot 

25°  C 

25°  C 

25°  C. 

Ratio  of  transformation                  .... 

2200  •  610 

2200  •  1500 

6400  •  2076 

No    load,    temp.  = 

Voltage  (line)  . 
Current  (line) 

2200 
4.14 

500 
36.7 

2200 
39.9 

2200 
75.1 

6400 
89.5 

25  C. 

Power  

1620 

7550 

13,600 

15,200 

47,500 

Voltage  (line)  . 

620 

80 

750 

400 

2200 

Blockedtemp.  =  25°C. 

Current  (line) 

25 

245 

288 

346.5 

369 

Power  

10,600 

12,500 

116,000 

68,000 

375,000 

Blocked  with  full  im- 

[Current (line)  . 

935 

1960 

1210 

pressed  voltage. 

Power,    (kw.)  

1050 

.  1960 

2950 

51.1  What  are  the  power  factor,  the  output,  and  the  efficiency 
of  the  induction  motor  No.  F  when  the  slip  is  1.55  per  cent. 
and  the  voltage  generated  by  the  air-gap  flux  in  the  stator  wind- 
ing is  2060  volts?  Assume  that  the  ratios  of  effective  to  ohmic 
resistance  are  respectively  1.1  and  1.20  for  the  stator  and  rotor 
windings  and  that  the  leakage  reactances  are  equal  when  reduced 
to  the  same  side. 

52. l  What  are  the  slip,  the  power  factor,  and  the  efficiency 
of  the  induction  motor  No.  G  when  it  delivers  full  load  and  the 


Use  the  transformer  diagram  in  the  solution  of  this  problem. 


84      PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

voltage  generated  in  the  stator  winding  by  the  air-gap  flux  is 
the  same  as  at  no  load?  Assume  that  the  ratios  of  effective  to 
ohmic  resistance  are  respectively  1.4  and  1.6  for  the  stator  and 
rotor  windings,  and  that  the  ratio  of  the  leakage  reactances 
at  the  rated  frequency  is  equal  to  the  ratio  of  the  effective 
resistances. 

53.  *  What  are  the  power  factor,  the  torque  and  the  efficiency 
of  the  induction  motor  No.  H  when  the  slip  is  1.86  per  cent.,  and 
the  voltage  generated  by  the  air-gap  flux  in  the  stator  winding 
is  93  per  cent,  of  the  rated  impressed  voltage?  Assume  that 
the  ratios  of  effective  to  ohmic  resistance  are  the  same  for  the 
stator  and  rotor  windings  at  25°  C.,  and  that  the  ratio  of  the 
ohmic  resistances  is  equal  to  the  ratio  of  the  leakage  reactances 
of  the  two  windings  at  the  rated  frequency.  The  temperature 
under  the  running  condition  is  65°  C. 

54. 1  What  are  the  power  factor,  the  slip,  and  the  efficiency 
of  the  induction  motor  No.  I  when  it  delivers  a  torque  of  22 , 000 
pound-feet,  and  the  voltage  generated  by  the  air-gap  flux  in 
the  stator  winding  is  2040  volts?  Assume  that  the  ratios  of 
the  ohmic  resistance,  the  effective  resistances,  and  the  leakage 
reactances  of  the  stator  and  rotor  windings  are  equal  at  the  rated 
frequency  and  a  temperature  of  25°  C.  The  running  tempera- 
ture is  70°  C. 

55. 1  What  are  the  power  factor,  the  slip,  and  the  efficiency 
when  the  induction  motor  No.  J  delivers  2750  h.p.  and  the 
voltage  generated  by  the  air-gap  flux  in  the  stator  winding  is 
92  per  cent,  of  the  rated  impressed  voltage?  Assume  that  the 
ratio  of  the  effective  to  the  ohmic  resistance  of  the  rotor  is 
20  per  cent,  greater  than  for  the  stator,  and  that  the  ratios 
of  the  leakage  reactances  and  of  the  effective  resistances  are 
equal  at  the  rated  frequency  and  a  temperature  of  25°  C.  The 
running  temperature  is  70°  C. 

56.  What  are  the  power  factor,  the  torque  and  the  efficiency 
for  the  induction  motor  No.  I  when  the  slip  is  1.6  per  cent, 
and  the  impressed  voltage  has  its  rated  value?     Make  the  same 
assumptions  in  regard  to  the  resistances  and  reactances  as  were 
made  in  problem  54. 

57.  What  are  the  power  factor,  the  slip  and  the  efficiency 
when  the  induction  motor  No.  J  delivers  a  torque  of  164,000 
pound-feet  and  the  impressed  voltage  is   6400  volts?     Make 

1  Use  the  transformer  diagram  in  the  solution  of  this  problem. 


INDUCTION  MOTORS  85 

the  same  assumptions  in  regard  to  the  resistances  and  reactances 
as  were  made  in  problem  55. 

58.  What  are  the  power  factor,  the  slip  and  the  efficiency  when 
the  induction  motor  No.  H  delivers  560  h.p.  and  the  impressed 
voltage  has  its  rated  value  ?     Make  the  same  assumption  in  regard 
to  the  resistances  and  reactances  as  were  made  in  problem  53. 

59.  Two  3-phase,  220-volt.  60-cycle  induction  motors  are  con- 
nected in  concatenation  across  a  220-volt  circuit.     Each  is  rated 
to  deliver  10  h.p.  and  is  wound  for  6  poles.     What  is  the  no-load 
speed?     When  they  deliver  10  h.p.  what  is  the  torque  developed 
by  each  motor? 

60.  Two  3-phase,  220-volt.  60-cycle  induction  motors  are  con- 
nected in  concatenation  across  a  220-volt  circuit.     Each  is  rated 
to  deliver  10  h.p.  but  one  iswound  for  6  poles  and  the  other  for 
8  poles.     What  is  the  no-load  speed?     When  they  deliver  10  h.p. 
what  is  the  torque  developed  by  each  motor?     What  per  cent, 
of  its  full-load  current  does  each  motor  carry? 

61.  Two  3-phase,  220-volt,  60-cycle  induction  motors  are  con- 
nected in  concatenation  across  a  220-volt  circuit.     One  is  rated 
to  deliver  10  h.p.  and  the  other  15  h.p.,  but  both  are  wound  for 
6  poles.     What  is  the  no-load  speed?     What  load  is  delivered 
when  the  15-h.p.  motor  takes  its  full-load  current?     What  per 
cent,   of  its  full-load  value  is  the  current  in  the  10-h.p.  motor? 
What  is  the  torque  developed  by  each? 

62.  Two  3-phase,  220  volt  60-cycle  induction  motors  are  con- 
nected in  concatenation  across  a  220-volt  circuit.     One  is  rated 
to  deliver  10  h.p.  and  is  wound  for  6  poles,  and  the  other  is  rated 
to  deliver  15  h.p.  and  is  wound  for  4  poles.     What  is  the  no-load 
speed?     What  is  the  greatest  load  that  can  be  delivered  and 
have  neither  motor  take  more  than  its  full-load  current?     When 
they  deliver  15  h.p.  what  torque  does  each  develop? 

63.  The  two  induction  motors,  M  and  P,  are  connected  in 
concatenation.     The  stator   of  the  first  receives  power  at  its 
rated  voltage  and  frequency,  and  the  stator  of  the  second  is 
short-circuited.     Neglect  the  no-load  component  of  the  current 
and  the  core  loss  due  to  the  leakage  flux. 

(a)  What  is  the  total  output  when  the  slip  of  the  first  motor 
is  51.3  per  cent.? 

(b)  What  is  the  power  developed  by  each  motor? 

(c)  What  are  the  copper  losses  in  the  stator  and  rotor  of  each 
motor? 


86     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

INDUCTION  MOTOR  DATA 


Number 

K 

L 

M 

N 

P 

Horse-power 

500 

570 

150 

150 

120 

Line  voltage  

2000 

1900 

500 

500 

500 

Connection  (both  stator  and  rotor)  

Y 

A 

Y 

Y 

Y 

Frequency 

60 

22.5 

38 

38 

38 

Poles  

44 

36 

6 

8 

6 

Magnetizing  current  (per  phasa.)  '  

23 

33 

45 

46 

34 

Core  loss  at  no  load1'2  

2470 

2800 

1300 

1360 

1200 

/  Stator  
Ohmic  resistance  per  phase    <  R 

0.24 
0.0153 

0.488 
0.0138 

0.054 
0.014 

0.052 
0.013 

0.04 
0.01 

Ratio  of  effective  resistance  f  Stator  

1.3 

1.35 

1.25 

1.25 

1.2 

to  ohmic  resistance  at  the  -j  Rotor  

1.4 

1.55 

1.4 

1.35 

1.4 

rated  frequency. 

Leakage   inductance  per  /  Stator  

2.6 

13. 

0.091 

0.086 

0.068 

phase  (mil-henry).            \  Rotor  

0.17 

0.37 

0.024 

0.022 

0.018 

Ratio  of  transformation  

16  to  5 

6  tol 

15  to  7 

9  to  5 

2  to  1 

64.  The  two  induction  motors,  N  and  P,  are  connected  in 
concatenation.     The  stator  of  the  first  receives  power  at  its 
rated  voltage  and  frequency,  and  the  stator  of  the  second  is 
short-circuited.     Neglect  the  no-load  component  of  the  current 
and  the  core  loss  due  to  the  leakage  flux.     Assume  that  the  core 
loss  due  to  the  mutual  flux  varies  as  the  product  of  the  frequency 
and  the  square  of  the  flux  density. 

(a)  What  is  the  total  output  when  the  slip  of  the  second  motor 
is  4.0  per  cent.? 

(b)  What  is  the  power  developed  by  each  motor? 

(c)  What  are  the  copper  losses  in  the  stator  and  rotor  of  each 
motor? 

(d)  What  are  the  core  losses  in  the  stator  and  rotor  of  each 
motor? 

65.  The  two  induction  motors,  M  and  N,  are  connected  in  con- 
catenation.    The  stator  of  the  first  receives  power  at  its  rated 
voltage  and  frequency,  and  the  stator  of  the  second  is  short- 
circuited.     Neglect  the  no-load  component  of  the  current.     As- 
sume that  the  loss  caused  by  the  leakage  flux  varies  as  the  fre- 
quency and  the  square  of  the  current.     Assume  that  the  core 
loss  due  to  the  mutual  flux  varies  as  the  product  of  the  frequency 
and  the  square  of  the  flux  density. 

(a)  What  is  the  total  output  when  the  speed  is  228  rev.  per  min.  ? 

1  At  the  rated  voltage. 

2  Assume  that  the  ratio  of  the  core  loss  in  the  stator  to  that  in  the  rotor 
for  the  same  mean  flux  density  and  frequency  is  4  to  3. 


INDUCTION  MOTORS  87 

(b)  What  is  the  power  developed  by  each  motor? 

(c)  What  are  the  effective  resistance  losses  in  the  stator  and 
rotor  of  each  motor? 

(d)  What  are  the  core  losses  in  the  stator  and  rotor  of  each 
motor? 

66.  The  two  induction  motors,   K  and  L,  are  connected  in 
concatenation.     The  stator  of  the  first  receives   power   at  its 
rated  voltage  and  frequency,  and  the  stator  of  the  second  is 
short-circuited.     Neglect  the  no-load  component  of  the  current. 
Assume  that  the  loss  caused  by  the  leakage  flux  is  proportional 
to  the  frequency  and  the  square  of  the  current.     Assume  that 
the  core  loss  due  to  the  mutual  flux  varies  as  the  frequency  and 
the  square  of  the  flux  density. 

(a)  What  is  the  total  output  when  the  slip  of  the  second  motor 
is  1.55  per  cent.? 

(b)  What  is  the  power  developed  by  each  motor? 

(c)  What  are  the  effective  resistance  losses  in  the  stator  and 
rotor  of  each  motor? 

(d)  What  are  the  core  losses  in  the  stator  and  rotor  of  each 
motor? 

67.  The  two  induction  motors,  M  and  P,  are  connected  in 
concatenation.     The  stator  of  the  first  receives  power  at  its 
rated  frequency,  and  the  stator  of  the  second  is  short-circuited. 
Do  not  neglect  the  no-load  component  of  the  current,  but  assume 
that  it  is  proportional  and  wattless  with  respect  to  the  generated 
voltage.     Assume  that  the  loss  caused  by  the  leakage  flux  varies 
as  the  frequency  and  the  square  of  the  current.     Assume  that 
the  core  loss  due  to  the  mutual  flux  varies  as  the  product  of  the 
frequency  and  the  square  of  the  flux  density. 

(a)  What  is  the  total  output  when  the  speed  is  268  rev.  per 
min.  and  the  voltage  generated  by  the  air-gap  flux  in  the  rotor 
winding  of  the  second  motor  is  68  volts  per  phase? 

(b)  What  power  do  they  receive  from  the  line  and  at  what 
power  factor  do  they  operate? 

(c)  What  is  the  power  developed  by  each  motor? 

68.  The  two  induction  motors,  K  and  L,  are  connected  in 
concatenation.     The  stator  of  the  first  receives  power  at  its 
rated  frequency,  and  the  stator  of  the  second  is  short-circuited. 
Do  not  neglect  the  no-load  component  of  the  currrent,  but  assume 
that  it  is  proportional  and  wattless  with  respect  to  the  generated 
voltage.     Assume  that  the  loss  caused  by  the  leakage  flux  varies 


88     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

as  the  frequency  and  the  square  of  the  current.  Assume  that 
the  core  loss  due  to  the  mutual  flux  varies  as  the  product  of  the 
frequency  and  the  square  of  the  flux  density. 

(a)  What  is  the  total  output  when  the  slip  of  the  second  is 
1.55  per  cent,  and  the  voltage  generated  by  the  air-gap  flux  in 
the  rotor  winding  of  the  second  motor  is  430  volts  per  phase? 

(b)  What  power  do  they  receive  from  the  line  and  at  what 
power  factor  do  they  operate? 

(c)  What  is  the  power  developed  by  each  motor? 

69.  A  1000-h.p.  induction  motor  is  operated  as  an  induction 
generator  in  parallel  with  a  synchronous  generator  having  a 
full-load  capacity  of  1000  kv.-a.     The  induction  machine,  which 
is  wound  for  12  poles,  is  driven  at  a  constant  speed  of  250.5  rev. 
per  min.     The  speed  of  the  synchronous  generator  falls   uni- 
formly from  1530  rev.  per  min.  at  no  load  to  1500  rev.  per  min. 
at  full  load,  when  the  frequency  is  25  cycles.     The  load  deliv- 
ered by  the  induction  generator  is  proportional  to  the  slip  which 
at  full  load  is  1.7  per  cent. 

When  the  total  load  supplied  is  1500  kw.  what  is  the  load 
delivered  by  each?  At  what  speed  should  the  induction  gener- 
ator be  driven  so  that  both  will  deliver  their  rated  loads  at  the 
same  time? 

70.  An  induction  generator  and  a  synchronous  generator,  each 
rated  to  deliver  2500  kv.-a.,  are  operated  in  parallel.     The  speed 
of  the  induction  generator  falls  from  1520  at  no  load  to  1498  at 
full    load,   and  the  speed  of  the  synchronous  generator  falls 
from  1525  at  no  load  to  1490  at  full  load.     The  load  on  the 
induction  generator  is  proportional  to  the  slip  which  at  full 
load  is  1.8  per  cent. 

What  is  the  greatest  load  that  .can  be  delivered  without  over 
loading  either  generator?  To  what  value  should  the  full-load 
speed  of  the  synchronous  generator  be  adjusted  so  that  both  gen- 
erators will  deliver  their  rated  loads  at  the  same  time? 

71.  A   4-pole  induction  generator  and  a  2-pole  synchronous 
generator  are  operating  in  parallel.     The  induction  generator  is 
driven  at  a  constant  speed,  but  the  speed  of  the  synchronous 
generator  falls  from  3660  rev.  per  min.  at  no  load  to  3590  at 
full  load  of  2000  kw.     The  resistance  of  the  rotor  windings 
between  terminals  refered  to  the  stator  is  0.73  ohm.     Neglect 
the^stator  resistance  and  reactance,  the  rotor  reactance,  and  the 
losses.     The  excitation  of  the  synchronous  generator  is  adjusted 


INDUCTION  MOTORS  89 

so  that  the  terminal  voltage  is  6400  volts  at  all  loads.  The 
speed  of  the  induction  generator  is  adjusted  so  that  when  the 
synchronous  machine  is  delivering  no  power  the  load  on  the 
induction  generator  is  500  kw.  What  is  the  division  of  the  load 
when  3500  kw.  is  required?  What  is  the  frequency  at  this 
time?  At  what  speed  should  the  induction  generator  be  driven 
so  that  both  will  deliver  their  rated  loads,  viz.,  2000  kw.,  at  the 
same  time? 

72.  A  500-kw.,  3-phase  induction  generator  is  operated  with 
a  synchronous  motor  floated  across  its  terminals.     At  no  load, 
when  running  as  an  induction  motor,  it  takes  a  line  current  of  31 
amperes  at  2000  volts.     The  resistance  of  the  rotor  winding  be- 
tween terminals  is  0.0228  ohm,  and  the  ratio  of  transformation 
from  stator  to  rotor  is  16  to  5.     Neglect  the  resistance  and 
reactance  of  the  stator  and  the  reactance  of  the  rotor  windings 
of  the  induction  generator,  and  all  of  the  losses  in  both  machines. 
The  excitation  of  the  synchronous  motor  is  adjusted  so  that  the 
terminal   voltage  is   2000   volts.     The  load  supplied  by  these 
machines  is  450  kw.  at  a  power  factor  of  85  per  cent.     What  is 
the  line  current  supplied  by  the  induction  generator?     What  is 
the   frequency   of   this   system   if    the   induction  generator   is 
wound  for  44  poles  and  is  driven  at  165  rev.  per  min.?     If  the 
synchronous  motor  has  a  synchronous  reactance  of  6.2  ohms 
between  terminals  what  is  its  necessary  excitation  voltage? 

73.  An  induction  generator  supplies  power  to  a  load  and  to 
an  over-excited  synchronous  motor.     At  no  load  as  an  induction 
motor  it  takes  a  line  current  of  98  amperes  at  2200  volts.     The 
resistance  of  the  rotor  winding  between  terminals  is  0.058  ohm 
and  the  ratio  of  transformation  from  stator  to  rotor  is  22  to  15. 
Neglect  the  resistance  and  reactance  of  the  stator  winding  and 
the  reactance  of  the  rotor  winding  of  the  induction  generator 
and  all  of  its  losses.     The  rotational  losses  of  the  synchronous 
motor  are  23.3  kw.  and  the  resistance  and  synchronous  reactance 
are  respectively  0.26  ohm  and   3.04  ohm   between  terminals. 
The  induction  generator  delivers   1150  kw.  of  which   the  syn- 
chronous motor  receives  450  kw.     The  excitation  of  the  latter 
is  adjusted  so  that  the  terminal  voltage  is  2200  volts.     The  power 
factor  of  the  load  exclusive  of  the  synchronous  motor  is  0.83. 
The  induction  generator  is  wound  for  12  poles  and  is  driven  at 
a  speed  of  254  rev.  per  min. 

What  is  the  frequency  of  the  system.     At  what  power  factor 


90     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

is  the  synchronous  motor  operating?  What  is  the  excitation 
voltage  of  the  synchronous  motor?  What  is  the  power  output 
of  the  synchronous^  motor? 

74.  A  2000-kw.  induction  generator  is  operated  in  parallel 
with  a  synchronous  generator  of  the  same  capacity.     At  no  load, 
when  running  as  an  induction  motor,  it  takes  a  line  current  of 
90  amperes  at  6400  volts.     The  resistance  of  the  rotor  winding 
between  terminals  is  0.0766  ohm  and  the  ratio  of  transformation 
from  stator  to  rotor  is  6400  to  2076.     Neglect  the  resistance 
and  reactance  of  the  stator  and  the  reactance  of  the  rotor  winding 
and  all  of  its  losses.     The  resistance  and  synchronous  reactance 
of  the  synchronous  generator  between  terminals  are  respectively 
0.65   ohm  and  14.2  ohms.     The  induction  generator   delivers 
1800  kw.  and  the  synchronous  generator  1200  kw.     The  excita- 
tion of  the  latter  is  adjusted  so  that  the  terminal  voltage  is  6400 
volts,  and  the  power  factor  of  the  load  is  0.85.     The  induction 
generator  is  wound  for  36  poles  and  is  driven  at  a  speed  of  82.4 
rev.  per  min. 

What  is  the  frequency  of  the  system?  At  what  power  factor 
does  the  synchronous  generator  operate?  What  is  the  excita- 
tion voltage  of  the  synchronous  generator? 

75.  An  induction  generator  and  a  synchronous  generator  are 
operated  in  parallel  and  supply  a  load  of  700  kw.  at  0.86  power 
factor.     Data  concerning  the  induction  generator  are:  370  kw., 
2000  volts,  44  poles,  Y  wound.     The  rotor  resistance  is  0.166 
ohm  per  phase  refered  to  the  stator.     The  magnetizing  current 
is  23  amperes  at  the  rated  voltage,  and  the  core  loss  is  2.47  kw. 
Neglect   the   resistance   and   reactance  of   the   stator  winding. 
Data  concerning  the  synchronous  generator  are:  500  kw.,  2200 
volts,  64  poles,  Y  wound.     The  effective  resistance  and  synchron- 
ous reactance  of  the  armature  winding  are  respectively  0.224 
ohm  and  2.76  ohms  per  phase.     The  induction  generator  is 
driven  at  166.4  rev.  per  min.  and   the  synchronous  generator 
at  113.0  rev.  per  min.     What  is  the  division  of  the  load  if  the 
excitation  of  the  synchronous  machine  is  adjusted  so  that  the 
terminal  voltage  is  2200  volts?    What  is  the  necessary  excitation 
voltage  of  the  synchronous  generator? 

76.  An  induction  generator  and  a  synchronous  generator  are 
operated  in  parallel  and  supply  a  load  of  950  kw.  at  0.83  power 
factor.     Data  concerning  the  induction  generator  are:  450  kw., 
2000  volts,  36  poles,  A  wound.     The  rotor  resistance  is  0.50  ohm 


INDUCTION  MOTORS  91 

per  phase  refered  to  the  stator.  The  magnetizing  current  is  33 
amperes  per  phase  and  the  core  loss  is  2.8  kw.  Neglect  the 
resistance  and  reactance  of  the  stator  winding.  Data  concern- 
ing the  synchronous  generator  are:  850  kw.,  2000  volts,  32  poles, 
Y  wound.  The  effective  resistance  and  synchronous  reactance 
are  respectively  0.064  ohm  and  2.40  ohms  per  phase.  The 
induction  generator  is  driven  at  85.0  rev.  per  min.  and  the  syn- 
chronous generator,  at  94.0  rev.  per  min.  What  is  the  division 
of  the  load  if  the  excitation  of  the  synchronous  generator  is 
adjusted  so  that  the  terminal  voltage  is  2000  volts?  What  is 
the  necessary  excitation  voltage  of  the  synchronous  generator? 

77. l  The  induction  motor  No.  G  is  operated  as  a  generator  on  a 
60-cycle  circuit  in  parallel  with  synchronous  apparatus.  What 
are  the  slip,  the  power  factor  and  the  efficiency  when  it  re- 
ceives 150  h.p.  from  the  prime  mover  and  the  voltage  generated 
in  the  stator  winding  by  the  air-gap  flux  is  460  volts?  Assume 
that  the  ratios  of  effective  to  ohmic  resistance  are  respectively 
1.4  and  1.6  for  the  stator  and  rotor  windings,  and  that  the 
ratio  of  the  leakage  reactances  at  the  rated  frequency  is  the  same 
as  ratio  of  the  effective  resistances. 

78.1  The  induction  motor  No.  H  is  operated  as  a  generator  on 
a  2200-volt,  25-cycle  circuit  in  paralled  with  synchronous  appa- 
ratus. What  are  the  output,  the  power  factor  and  the  efficiency 
when  the  slip  is  1.86  per  cent,  and  the  voltage  generated  in  the 
stator  winding  by  the  air-gap  flux  is  assumed  to  be  93  per  cent, 
of  the  terminal  voltage?  Assume  that  the  ratios  of  effective  to 
ohmic  resistance  are  the  same  at  25°  Centigrade,  and  that  the 
ratio  of  the  ohmic  resistances  is  equal  to  the  ratio  of  the  leakage 
reactances  of  the  two  windings  at  the  rated  frequency.  The 
temperature  under  the  running  condition  is  65°  C. 

79. 1  The  induction  motor  No.  I  is  operated  as  a  generator 
on  a  25-cycle  circuit  in  parallel  with  syncrohnous  apparatus. 
What  are  the  slip,  the  power  factor  and  the  efficiency  when  it  re- 
ceives 1000  h.p.  from  the  prime  mover  and  the  voltage  generated 
in  the  stator  winding  by  the  air-gap  flux  is  2040  volts?  Assume 
that  the  ratios  of  the  ohmic  resistances,  the  effective  resistances 
and  the  leakage  reactances  of  the  stator  and  rotor  windings  are 
equal  at  the  rated  frequency  and  a  temperature  of  25°  C. 
The  running  temperature  is  70°  C. 

80. 1  The  induction  motor  No.  /  is  operated  as  a  generator 

*See  page  83. 


92     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

on  a  6400-volt,  25-cycle  circuit  in  parallel  with  synchronous 
apparatus.  What  are  the  output,  the  power  factor,  and  the 
efficiency  when  the  slip  is  1.9  per  cent,  and  the  voltage  gener- 
ated in  the  stator  winding  by  the  air-gap  flux  is  (assumed  to  be) 
92  per  cent,  of  the  terminal  voltage?  Assume  that  the  ratio  of 
the  effective  to  the  ohmic  resistance  for  the  rotor  is  20  per  cent, 
greater  than  for  the  stator,  and  that  the  ratios  of  the  leakage 
reactances  and  of  the  effective  resistances  are  equal  at  the  rated 
frequency  and  a  temperature  of  25°  C.  The  running  temperature 
is  70°  C. 


CHAPTER  V 
ROTARY  CONVERTERS 

1.  Assume  that  the  graph  representing  the  flux  density  in  the 
air-gap  of  a  rotary  converter  is  rectangular  and  is  constant  over 
the  entire  pole  pitch.     Also  assume  that  the  armature  winding 
is  uniformly  distributed.     Calculate  the  ratio     of  the  single- 
phase  alternating-current  voltage  to  the  direct-current  voltage. 
Assume  that  the  coil  pitch  and  phase  spread  are  each  unity. 

2.  In  problem  1  calculate  the  ratio  of  the  four-phase  alternat- 
ing-current voltage  to  the  direct-current  voltage.     Assume  that 
the  coil  pitch  is  one  and  that  the  phase  spread  is  one-half. 

3.  In  problem  1  calculate  the  ratio  of  the  three-phase  alternat- 
ing-current voltage  to  the  direct-current  voltage.     Assume  that 
the  coil  pitch  is  one  and  that  the  phase  spread  is  two-thirds. 

4.  In  problem  1  calculate  the  ratio  of  the  six-phase  alternating- 
current  voltage  to  the  direct-current  voltage.     Assume  that  the 
coil  pitch  is  one  and  that  the  phase  spread  is  one- third. 

5.  In  problem  1  calculate  the  ratio  of  the  twelve-phase  alternat- 
ing-current voltage  to  the  direct-current  voltage.     Assume  that 
the  coil  pitch  is  one  and  that  the  phase  spread  is  one-sixth. 

6.  Assume  that  the  air-gap  flux  density  in  a  rotary  converter 
is  constant  under  the  poles  and  is  zero  between  them.     The 
ratio  of  pole  arc  to  pole  pitch  is  two-thirds.     Also  assume  that 
the  armature  winding  is  uniformly  distributed.     Calculate  the 
ratio   of   the    single-phase    alternating-current    voltage   to   the 
direct-current  voltage.     Assume  that  the  coil  pitch  and  the  phase 
spread  are  each  unity. 

7.  In  problem  6  calculate  the  ratio  of  the  four-phase  alternat- 
ing-current voltage  to  the  direct-current  voltage.     Assume  that 
the  coil  pitch  is  one  and  that  the  phase  spread  is  one-half. 

8.  In  problem  6  calculate  the  ratio  of  the  three-phase  alternat- 
ing-current voltage  to  the  direct-current  voltage.     Assume  that 
the  coil  pitch  is  one  and  that  the  phase  spread  is  two-thirds. 

9.  In  problem  6  calculate  the  ratio  of  the  six-phase  alternating- 


94     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

current  voltage  to  the  direct-current  voltage.     Assume  that  the 
coil  pitch  is  one  and  that  the  phase  spread  is  one-third. 

10.  In  problem  6  calculate  the  ratio  of  the  twelve-phase  alter- 
nating-current voltage  to  the  direct-current  voltage.     Assume 
that  the  coil  pitch  is  one  and  that  the  phase  spread  is  one-sixth. 

11.  The  graph  representing  the  flux  density  in  the  air-gap  of 
a  rotary  converter  is  a  simple  harmonic  function.     Assume  that 
the  armature  winding  is  uniformly  distributed.     Calculate  the 
ratio    of   the    single-phase    alternating-current   voltage    to    the 
direct-current  voltage.     Assume   that   the   coil  pitch   and  the 
phase  spread  are  each  unity. 

12.  In  problem  11  calculate  the  ratio  of  the  four-phase  alter- 
nating-current voltage  to  the  direct-current  voltage.     Assume 
that  the  coil  pitch  is  one  and  that  the  phase  spread  is  one-half. 

13.  In  problem  11  calculate  the  ratio  of  the  three-phase  alter- 
nating-current voltage  to  the  direct-current  voltage.     Assume 
that  the  coil  pitch  is  one  and  that  the  phase  spread  is  two-thirds. 

14.  In  problem  11  calculate  the  ratio  of  the  six-phase  alter- 
nating-current voltage  to  the  direct-current  voltage.     Assume 
that  the  coil  pitch  is  one  and  that  the  phase  spread  is  one-third. 

15.  In  problem  1 1  calculate  the  ratio  of  the  twelve-phase  alter- 
nating-current voltage  to  the  direct-current  voltage.     Assume 
that  the  coil  pitch  is  one  and  that  the  phase  spread  is  one-sixth. 

16.  The  graph  representing  the  flux  distribution  in  the  air- 
gap  of  a  rotary  converter  is  B  =  Bisin.x  +  B3sin  3x.     x  is  the 
angular  displacement  measured  from  the  neutral  point.     Take 
the  third  harmonic  component  of  the  flux  density  as  0.3  of  the 
fundamental.     Assume  that  the  armature  winding  is  uniformly 
distributed.     Calculate  the  ratio  of  the  single-phase  alternating- 
current  voltage  to  the  direct-current  voltage.     Assume  that  the 
coil  pitch  and  the  phase  spread  are  each  unity. 

17.  In  problem  16  calculate  the  ratio  of  the  four-phase  alter- 
nating-current voltage  to  the  direct-current  voltage.     Assume 
that  the  coil  pitch  is  one  and  that  the  phase  spread  is  one-half. 

18.  In  problem  16  calculate  the  ratio  of  the  three-phase  alter- 
nating-current voltage  to  the  direct-current  voltage.     Assume 
that  the  coil  pitch  is  one  and  that  the  phase  spread  is  two-thirds. 

19.  In  problem  16  calculate  the  ratio  of  the  six-phase  alter- 
nating-current voltage  to  the  direct-current  voltage.     Assume 
that  the  coil  pitch  is  one  and  that  the  phase  spread  is  one-third. 

20.  In  problem   16  calculate  the  ratio  of  the  twelve-phase 


ROTARY  CONVERTERS  95 

alternating-current  voltage  to  the  direct-current  voltage.  As- 
sume that  the  coil  pitch  is  one  and  that  the  phase  spread  is 
one-sixth. 

21.  The  graph  representing  the  flux  distribution  in  the  air- 
gap   of  a  rotary  converter  is  B  =  Bisin.x  —  .Bssin  3x.     x  is  the 
angular  displacement  measured  from  the  neutral  point.     Take 
the  third  harmonic  component  of  the  flux  density  as  0.3  of  the 
fundamental.     Assume  that  the  armature  winding  is  uniformly 
distributed.     Calculate  the  ratio  of  the  single-phase  alternating- 
current  voltage  to  the  direct-current  voltage.     Assume  that  the 
coil  pitch  and  the  phase  spread  are  each  unity. 

22.  In  problem  21  calculate  the  ratio  of  the  four-phase  alter- 
nating-current voltage  to  the  direct-current  voltage.     Assume 
that  the  coil  pitch  is  one  and  that  the  phase  spread  is  one-half. 

23.  In  problem   21    calculate   the  ratio   of  the   three-phase 
alternating-current  voltage  to  the  direct-current  voltage.     As- 
sume that  the  coil  pitch  is  one  and  that  the  phase  spread  is 
two-thirds. 

24.  In  problem  21  calculate  the  ratio  of  the  six-phase  alter- 
nating-current voltage  to  the  direct-current  voltage.     Assume 
that  the  coil  pitch  is  one  and  that  the  phase  spread  is  one-third. 

25.  In  problem  21   calculate  the  ratio  of  the  twelve-phase 
alternating-current  voltage  to  the  direct-current  voltage.     As- 
sume that  the  coil  pitch  is  one  and  that  the  phase  spread  is  one- 
sixth. 

26.  In  a  single-phase  rotary  converter,  assume  that  the  cur- 
rents on  the  direct-  and  alternating- current  sides  are  respectively 
steady  and  sinusoidal,  and  neglect  all  of  the  losses  in  calculating 
their  relative  values,     (a)  Calculate  the  ratio  of  the  average, 
heating  in  a  conductor  at  one  of  the  alternating-current  taps  to 
that  in  a  conductor  midway  between  the  taps  when  the  rotary 
is   operating   at   unit  power  factor,     (b)  Calculate   this  ratio 
when  the  rotary  is  operating  at  0.7  power  factor. 

27.  In  a  four-phase  rotary  converter  assume  that  the  currents 
on  the  direct-  and  alternating-current  sides  are  respectively 
steady  and  sinusoidal,  and  neglect  all  of  the  losses  in  calculating 
their  relative  values,     (a)  Calculate  the  ratio  of  the  average 
heating  in  a  conductor  at  one  of  the  alternating-current  taps  to 
that  in  a  conductor  midway  between  the  taps  when  the  rotary 
is   operating   at   unit  power  factor,     (b)  Calculate   this   ratio 
when  the  rotary  is  operating  at  0.7  power  factor. 


96     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

28.  In  a  three-phase  rotary  converter  assume  that  the  cur- 
rents on  the  direct-  and  alternating-current  sides  are  respec- 
tively steady  and  sinusoidal,  and  neglect  all  of  the  losses  in 
calculating  their  relative  values,     (a)  Calculate  the  ratio  of  the 
average  heating  in  a  conductor  at  one  of  the  alternating-current 
taps  to  that  in  a  conductor  midway  between  the  taps  when  the 
rotary  is  operating  at  unit  power  factor,     (b)   Calculate  this 
ratio  when  the  rotary  is  operating  at  0.7  power  factor. 

29.  In  a  six-phase  rotary  converter  assume  that  the  currents 
on   the  direct-   and   alternating-current   sides   are   respectively 
steady  and  sinusoidal,  and  neglect  all  of  the  losses  in  calculating 
their  relative  values,     (a)  Calculate  the  ratio  of  the  average 
heating  in  a  conductor  at  one  of  the  alternating-current  taps  to 
that  in  a  conductor  midway  between  the  taps  when  the  rotary 
is   operating   at   unit  power   factor,     (b)  Calculate   this   ratio 
when  the  rotary  is  operating  at  0.7  power  factor. 

30.  In  a  twelve-phase  rotary  converter  assume  that  the  cur- 
rents on  the  direct-  and  alternating-current  sides  are  respec- 
tively steady  and  sinusoidal,  and  neglect  all  of  the  losses  in 
calculating  their  .relative  values,     (a)  Calculate  the  ratio  of  the 
average  heating  in  a  conductor  at  one  of  the  alternating-current 
taps  to  that  in  a  conductor  midway  between  the  taps  when  the 
rotary  is  operating  at  unit  power  factor,     (b)  Calculate  this 
ratio  when  the  rotary  is  operating  at  0.7  power  factor. 

31.  In  a  single-phase  rotary  converter  assume  that  the  cur- 
rents on  the  direct-  and  alternating-current  sides  are  respec- 
tively steady  and  sinusoidal  and  neglect  all  of  the  losses  in  cal- 
culating their  relative  values,     (a)  Calculate  the  relative  out- 
puts when  operating  as  a  rotary  converter  at  unit  power  factor 
and  as  a  direct-current  generator  on  the  basis  of  the  same  arma- 
ture copper  loss,     (b)  Calculate  the  relative  outputs  when  the 
rotary  is  operating  at  0.7  power  factor,     (c)  Calculate  the  rela- 
tive outputs  when  operating  as  a  rotary  converter  and  as  a 
synchronous  generator  at  unit  power  factor  on  the  basis  of  the 
same  armature  copper  loss,     (d)  Calculate  the  relative  outputs 
when  both  are  operating  at  0.7  power  factor. 

32.  In  a  four-phase  rotary  converter  assume  that  the  currents 
on  the  direct-  and  alternating-current  sides  are  respectively  steady 
and  sinusoidal,  and  neglect  all  of  the  losses  in  calculating  their 
relative  values,     (a)  Calculate  the  relative  outputs  when  operat- 
ing as  a  rotary  converter  at  unit  power  factor  and  as  a  direct- 


ROTARY  CONVERTERS  97 

current  generator  on  the  basis  of  the  same  armature  copper  loss. 
(b)  Calculate  the  relative  outputs  when  the  rotary  is  operating 
at  0.7  power  factor,  (c)  Calculate  the  relative  outputs  when 
operating  as  a  rotary  converter  and  as  a  synchronous  generator 
at  unit  power  factor  on  the  basis  of  the  same  armature  copper 
loss,  (d)  Calculate  the  relative  outputs  when  both  are  operat- 
ing at  0.7  power  factor. 

33.  In  a  three-phase  rotary  converter  assume  that  the  cur- 
rents on  the  direct-  and  alternating-current  sides  are  respectively 
steady  and  sinusoidal,  and  neglect  all  of  the  losses  in  calculating 
their  relative  values,     (a)  Calculate  the  relative  outputs  when 
operating  as  a  rotary  converter  at  unit  power  factor  and  as  a 
direct-current  generator   on   the  basis  of  the  same   armature 
copper    loss,     (b)  Calculate    the    relative    outputs    when    the 
rotary  is  operating  at  0.7  power  factor,     (c)  Calculate  the  rela- 
tive outputs  when  operating  as  a  rotary  converter  and  as  a  syn- 
chronous generator  at  unit  power  factor  on  the  basis  of  the 
same  armature  copper  loss,     (d)  Calculate  the  relative  outputs 
when  both  are  operating  at  0.7  power  factor. 

34.  In  a  six-phase  rotary  converter  assume  that  the  currents 
on  the   direct-   and   alternating-current   sides   are   respectively 
steady  and  sinusoidal,  and  neglect  all  of  the  losses  in  calculating 
their  relative  values,     (a)  Calculate  the  relative  outputs  when 
operating  as  a  rotary  converter  at  unit  power  factor  and  as  a 
direct-current  generator  on  the  basis  of  the  same  armature  copper 
loss,     (b)  Calculate  the  relative  outputs  when  the  rotary  is 
operating  at  0.7  power  factor,     (c)  Calculate  the  relative  out- 
puts when  operating  as  a  rotary  converter  and  as  a  synchronous 
generator  at  unit  power  factor  on  the  basis  of  the  same  armature 
copper  loss,     (d)  Calculate  the  relative  outputs  when  both  are 
operating  at  0.7  power  factor. 

35.  In  a    twelve-phase    rotary    converter  assume    that    the 
currents  on  the  direct-  and  alternating-current  sides  are  respec- 
tively steady  and  sinusoidal  and  neglect  all  of  the  losses  in  cal- 
culating their  relative  values,     (a)  Calculate  the  relative   out- 
puts when  operating  as  a  rotary  converter  at  unit  power  factor 
and  as  a  direct-current  generator  on  the  basis  of  the  same  armature 
copper  loss,     (b)  Calculate  the  relative  outputs  when  the  rotary 
is  operating  at  0.7  power  factor,     (c)  Calculate  the  relative  out- 
puts when  operating  as  a  rotary  converter  and  as  a  synchronous 
generator  at  unit  power  factor  on  the  basis  of  the  same  armature 


98     PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

copper  loss,     (d)  Calculate  the  relative  outputs  when  both  are 
operating  at  0.7  power  factor. 

36.  A  5-kilowatt  single-phase  rotary  converter  supplies  power 
on  the  direct-current  side  at  110  volts.     It  receives  energy  on  the 
alternating-current  side  through  a  transformer  from  a  2200-volt 
circuit.     What  should  be  the  transformer's  ratio  of  transforma- 
tion?    If  the  rotary  has  an  efficiency  of  85  per  cent,  what  should 
be  the  current  rating  of  the  high-tension  winding  of  the  trans- 
former? 

37.  A  100-kilowatt,  4-phase  rotary  converter  supplies  power 
on  the  direct-current  side  at  230  volts.     It  receives  energy  on 
the   alternating-current   side   through   two   single-phase   trans- 
formers from  a  2-phase  circuit.     The  voltage  between  adjacent 
high-tension  conductors  is  1555  volts,  and  between   alternate 
conductors  is  2200  volts.     The  high-tension  windings  of  the 
transformers  are  connected  across  the  2200-volt  lines  and  the 
low-tension  windings  are  connected  in  star  with  the  neutral 
point  grounded.     The  efficiency  of  the  rotary  at  full  load  and 
unit  power  factor  is  94  per  cent.     What  should  be  the  full-load 
current  and  voltage  ratings  of  the  high-  and  low-tension  windings 
of  the  transformers? 

38.  A  4-phase,  50-kilowatt  rotary  converter  supplies  power  on 
the  direct-current  side  at  220  volts.     It  receives  energy  on  the 
alternating-current   side    from    a    2-phase    line    through    two 
single-phase    transformers    which    have    double    primary  and 
secondary  windings.     The  voltage  between  the  adjacent  high- 
tension  conductors  is  1625  volts  and  between  alternate  conduc- 
tors is  2300  volts.     The  high-tension  windings  of  the  transform- 
ers are  connected  in  star  and  the  low-tension  windings  in  mesh. 
The  efficiency  of  the  rotary  at  full-load  output  and  0.90  power 
factor  is  90.2  per  cent.     What  are  the  primary  and  secondary 
currents  and  voltages  when  the  rotary  delivers  its  rated  load  at 
0.90  power  factor? 

39.  A  4-phase  100-kilowatt  rotary  converter  delivers  power 
on  the  direct-current  side  at  220  volts.     It  receives  energy  from 
a  three-phase,  11,000-volt  circuit  through  Scott-connected  trans- 
formers.    The  low-tension  windings  are  connected  in  star.     The 
efficiency  of  the  rotary  at  full  load  and  unit  power  factor  is  92.5 
per  cent.  What  are  the  ratios  of  transformation  of  each  of  the 
transformers?     What  are  the  full-load  current  ratings  of  the 
primary  and  secondary  windings  of  each  transformer? 


ROTARY  CONVERTERS  99 

40.  A  4-phase,  100-kilowatt,  230-volt  rotary  converter  receives 
its  power  from  a  three-phase,  6600-volt  circuit  through  Scott-con- 
nected transformers.     The  low-tension  windings  are  double  and 
are  connected  in  mesh.     The  efficiency  of  the  rotary  at  full  load 
and  0.90  power  factor  is  91.8  per  cent.     What  are  the  high-  and 
low-tension  currents  and  voltages  of  the  transformers  when  the 
rotary  delivers  its  rated  load  and  operates  at  0.90  power  factor? 

41.  A   three-phase,    200-kilowatt   rotary    converter    supplies 
direct  current  at  110  volts.     It  receives  power  from  a  three-phase, 
6600-volt  circuit  through  three  single-phase  transformers  which 
have  their  high-tension  windings  connected  in  delta  and  their  low- 
tension  windings  connected  in  Y.     At  full  load  and  unit  power 
factor  the  rotary  has  an  efficiency  of  93  per  cent.     What  should 
be  the  full-load  current  and  voltage  ratings  of  the  high  -and  low- 
tension  windings  of  the  transformers? 

42.  A   three-phase,   250-kilowatt,   220-volt  rotary   converter 
receives  power  through  three  single-phase  transformers  from  a 
3-phase,  2200-volt  circuit.     The  high-tension  windings  of  the 
transformers  are  connected  in  Y  and  the  low-tension  windings  in 
delta.     The  rotary  has  an  overload  current  capacity  of  15  per 
cent,  when  operating  at  0.9  power  factor  and  at  this  time  the 
efficiency  is  91.8  per  cent.     What  should  be  the  overload  current 
and  voltage  ratings  of  the  high-  and  low-tension  windings  of  the 
transformers? 

43.  A  three-phase,   300-kilowatt,   500-volt    rotary   converter 
receives  power  through  three  single-phase  transformers  from  a 
three-phase,  11,000-volt  circuit.     Both  the  high-  and  low-tension 
windings  of  the  transformers  are  connected  in  Y.     When  deliver- 
ing its  rated  load  and  operating  at  0.85  power  factor  the  rotary 
has  an  efficiency  of  91.6  per  cent.     What  is  the  ratio  of  trans- 
formation of  the  transformers?     What  should  be  the  full-load 
current  ratings  of  the  high-  and  low-tension  windings  of  the 
transformers? 

44.  A   three-phase,    150-kilowatt,    230-volt   rotary    converter 
receives  power  through  two  single-phase  transformers  from  a 
three-phase,  2200-volt  circuit.     Both  the  primary  and  secondary 
windings  of  the.  transformers  are  connected  in  V.     Neglecting 
the  losses  what  should  be  the  current  and  voltage  ratings  of  the 
high-  and  low- tension  windings  of  the  transformers? 

45.  A  6-phase,   1500-kilowatt,  650-volt  rotary  converter  de- 
livers power  to  a  railway  system.     It  receives  power  through 


100  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

three  single-phase  transformers  from  a  three-phase,  13,200- 
volt  transmission  line.  The  transformers  are  connected  in  Y  on 
the  high-tension  side  and  diametrically  on  the  low-tension  side. 
In  order  to  maintain  a  constant  line  voltage  of  13,200  volts  at  the 
substation  the  rotary  is  over-compounded  so  that  when  delivering 
15  per  cent,  overload  current  at  650  volts  it  operates  at  0.85  power 
factor  with  an  efficiency  of  92.7  per  cent.  What  is  the  ratio  of 
transformation  for  each  transformer?  What  should  be  the 
overload  current  ratings  of  the  high-  and  low-tension  windings  of 
the  transformers? 

46.  A  six-phase,    1000-kw.,  rotary  converter  receives  power 
through  three  transformers  from  a  three-phase,  11,000-volt  circuit. 
The  rotary  delivers  power  at  650  volts  to  a  railway  system.     The 
transformers  are  connected  in  Y  on  the  high-tension  and  in  double 
delta  on  the  low-tension  side.     What  should  be  the  current  and 
voltage  ratings  and  the  ratio  of  transformation  for  each  of  these 
transformers?     In  determining  the  rating  neglect  the  losses. 

47.  A  6-phase,  2000-kw.,  rotary  converter  receives  power  from 
a  3-phase,  13,000-volt  transmission  line.     The  transformers  are 
connected  in  delta  on  the  high-tension  and  in  double  Y  on  the 
low-tension  side.     The  rotary  delivers  power  at  650  volts  to  a 
railway  system.     It  has  an  overload  current  capacity  of  20  per 
cent,  and  an  efficiency  of  92.8  per  cent,  when  delivering  this  load 
and  operating  at  0.85  power  factor.     What  should  be  the  over- 
load current  and  voltage  ratings  of  the  high-  and  low-tension 
windings  of  the  transformers?     What  should  be  their  ratio  of 
transformation? 

48.  A  12-phase,  3000-kw.,  650-volt  rotating  converter  receives 
power  from  a  three-phase  transmission  line  through  three  single- 
phase  transformers,  whose  high-tension  windings  are  connected 
in  Y  and  whose  low-tension  windings  are  connected  in  double 
chord.     The  high-tension  line  voltage  is  22,000  volts.     Neglect- 
ing the  losses  what  should  be  the  current  and  voltage  ratings  of 
the  high-  and  low-tension  windings  of  the  transformers? 

49. l  A  6-phase,  25-cycle,  600-volt  rotary  converter  has  an 
efficiency  of  92.5  per  cent,  when  delivering  750  kw.  and  operating 
at  0.90  power  factor.  The  armature  winding  has  4  inductors 

1  In  calculating  the  armature  reaction  do  not  neglect  the  distribution  of 
the  winding,  and  assume  that  the  constant  usually  given  as  0.707  is  0.60. 
This  makes  an  approximate  correction  for  the  effect  of  the  ratio  of  pole  arc 
to  pole  pitch. 


ROTARY  CONVERTERS  101 

in  series  per  slot  and  24  slots  per  pole.  Neglect  the  resistance 
and  the  leakage  reactance  of  the  armature  winding.  What  are 
the  demagnetizing  ampere  turns  per  pole  for  the  Speerfibd' 
load? 

50. 1  A  3-phase,  25-cycle  rotary  converter  has  a /^idl-lp'acV 
capacity  of  300  kw.  at  600  volts.  The  armature  has  96  slots 
with  6  inductors  in  series  per  slot.  The  field  structure  has  4 
poles.  Neglect  the  resistance  and  leakage  reactance  of  the  arma- 
ture winding.  When  this  rotary  receives  a  line  current  of  540 
amperes  at  367  volts  and  is  overexcited  so  that  it  operates  at  a 
power  factor  of  0.90  what  are  the  demagnetizing  ampere  turns 
per  pole? 

51. l  A  6-phase,  60-cycle,  600-volt  rotary  converter  has  a  full- 
load  capacity  of  1000  kw.  when  operating  at  unit  power  factor. 
The  armature  has  180  slots  with  6  inductors  in  series  per  slot. 
The  field  structure  has  12  poles  each  of  which  is  wound  with  864 
turns.  When  this  rotary  is  overexcited  so  that  it  operates  at 
0.95  power  factor  and  is  delivering  its  rated  output  on  the  direct- 
current  side  the  efficiency  is  92.9  per  cent.  How  much  greater 
is  the  field  current  than  its  normal  value,  i.e.,  if  the  power  factor 
were  unity? 

52. l  At  full  load  and  when  operating  at  unit  power  factor  a 
3-phase,  25-cycle  rotary  converter  receives  a  line  current  of  499 
amperes  at  367  volts  and  delivers  on  the  direct-current  side  500 
amperes  at  600  volts.  The  armature  has  96  slots  with  6  inductors 
in  series  per  slot.  The  field  structure  has  4  poles  with  2340  turns 
in  the  shunt  windings  per  pole.  The  resistance  of  this  field 
circuit  with  the  regulating  rheostat  cut  out  is  120.5  ohms.  When 
this  rotary  receives  its  rated  current  what  is  the  least  power 
factor  at  which  it  can  be  operated  and  still  maintain  its  rated 
voltage?  Normal  excitation,  i.e.,  when  operating  at  unit 
power  factor,  is  3.25  amperes. 

53. l  At  full  load  and  when  operating  at  unit  power  factor  a  6- 
phase,  25-cycle  compound  rotary  converter  receives  a  line  current 
of  840  amperes  at  a  voltage  of  212  volts  between  adjacent  slip 
rings  and  delivers  on  the  direct-current  side  1667  amperes  at  600 
volts.  The  armature  has  168  slots  with  6  inductors  in  series  per 

1  In  calculating  the  armature  reaction  do  not  neglect  the  distribution  of 
the  winding,  and  assume  that  the  constant  usually  given  as  0.707  is  0.60. 
This  makes  an  approximate  correction  for  the  effect  of  the  ratio  of  pole  arc 
to  pole  pitch. 


102  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

slot.  The  field  structure  has  8  poles  with  1501  turns  in  the  shunt 
winding  and  2  turns  in  the  series  winding  per  pole.  A  current  of 
.56  amperes  in  the  shunt  field  winding  alone  is  the  normal  excita- 
tion and  produces  a  voltage  of  600  volts  on  the  direct-current 
.side  at  no  load  when  running  as  a  generator. 

Whep"  this  rotary  delivers  1500  amperes  at  600  volts  what  should 
be  the  shunt  field  current  in  order  that  the  current  on  the  alter- 
nating-current side  shall  be  leading  and  not  exceed  its  full-load 
value  and  the  power  factor  have  its  least  value?  What  is  this 
power  factor?  Neglect  any  change  in  the  efficiency. 

54. l  At  full  load  and  when  operating  at  unit  power  factor  a 
6-phase,  60-cycle  rotary  converter  receives  a  line  current  of  840 
amperes  at  a  voltage  of  424  volts  between  diametrical  points  of 
the  armature  and  delivers  on  the  direct-current  side  a  current  of 
1667  amperes  at  600  volts.  The  armature  has  15  slots  per  pole 
with  6  inductors  in  series  per  slot.  On  each  field  pole  there  are 
864  turns  in  the  shunt  winding  and  2  turns  in  the  series  winding. 
A  current  of  9.3  amperes  in  the  shunt  field  winding  alone  is  the 
normal  excitation  and  produces  a  voltage  of  600  volts  on  the 
direct-current  side  at  no  load  when  running  as  a  generator. 
When  this  rotary  delivers  1460  amperes  at  600  volts  what  are 
the  limits  of  the  shunt  field  current  in  order  that  the  current 
on  the  alternating-current  side  shall  not  exceed  its  full-load 
value  and  the  power  factor  have  its  least  value?  What  are 
these  limiting  power  factors?  Neglect  any  change  in  the 
efficiency. 

55. l  At  full  load  and  when  operating  at  unit  power  factor  a 
6-phase,  25-cycle  rotary  converter  takes  a  line  current  of  840 
amperes  at  a  voltage  of  424  between  diametrical  points  and 
delivers  1667  amperes  at  600  volts.  The  armature  has  21  slots 
per  pole  with  6  inductors  in  series  per  slot.  Each  field  pole  is 
wound  with  1501  turns.  The  resistance  of  the  field  circuit  with 
the  regulating  rheostat  cut  out  is  76.6  ohms.  A  current  of  5.6 
amperes  in  the  field  winding  is  the  normal  excitation:  i.e.,  for 
full  load  and  unit  power  factor.  With  an  output  of  850  kw.  at 
600  volts  what  is  the  least  power  factor  at  which  this  rotary  can 
be  operated  when  overexcited?  What  per  cent,  is  the  current 

1  In  calculating  the  armature  reaction  do  not  neglect  the  distribution  of 
the  winding,  and  assume  that  the  constant  usually  given  as  0.707  is  0.60. 
This  makes  an  approximate  correction  for  the  effect  of  the  ratio  of  pole  arc 
to  pole  pitch. 


ROTARY  CONVERTERS  103 

on  the  alternating-current  side  of  its  full-load  value?  Neglect 
any  change  in  the  efficiency. 

56. J  At  full  load  and  when  operating  at  unit  power  factor  a 
6-phase,  60-cycle  rotary  converter  takes  a  line  current  of  840 
amperes  at  212  volts  per  phase  and  delivers  1000  kw.  at  600 
volts.  The  armature  has  180  slots  with  6  inductors  in  series 
per  slot.  The  field  structure  has  12  poles  with  864  turns  in  the 
shunt  winding  and  2  turns  in  the  series  winding  per  pole.  The 
resistance  of  the  shunt  field  circuit  is  39.65  ohms  with  the  regu- 
lating rheostat  cut  out.  A  current  of  9.3  amperes  in  the  shunt 
field  alone  is  the  normal  excitation,  i.e.,  for  full  load  and  unit 
power  factor.  With  an  output  of  800  kw.  at  600  volts  what  is 
the  least  power  factor  at  which  the  rotary  can  be  operated  when 
overexcited?  What  per  cent,  is  the  current  of  its  full-load  value? 
Neglect  any  change  in  this  efficiency. 

57. l  A  3-phase,  25-cycle  rotary  converter  has  a  full-load  capa- 
city of  300  kw.  at  600  volts.  The  armature  has  96  slots  with 
6  inductors  in  series  per  slot.  The  field  structure  has  4  poles 
with  2340  turns  in  the  shunt  winding  per  pole.  The  resistance 
of  this  field  circuit  with  all  of  the  regulating  rheostat  cut  out  is 
120.5  ohms.  A  current  of  3.25  amperes  in  the  shunt  field  winding 
alone  gives  the  normal  excitation,  i.e.,  for  full  load  at  unit  power 
factor.  In  order  to  maintain  the  terminal  voltage  on  the  direct- 
current  side  at  600  volts  when  full  load  is  delivered,  it  is  neces- 
sary to  overexcite  the  rotary  so  that  it  takes  a  line  current  of 
600  amperes  at  a  power  factor  of  86  per  cent. 

Calculate  the  least  number  of  series  field  turns  that  must  be 
added  in  order  that  this  excitation  may  be  produced.  What 
should  be  the  current  in  the  shunt  field  winding? 

58. l  Calculate  the  efficiency  of  the  rotary  A  when  it  receives  300 
kw.  at  a  power  factor  of  0.90  and  with  a  line  voltage  of  380 
volts.  The  excitation  is  greater  than  normal.  Assume  that 
the  running  temperature  is  70°  C. 

59. 1  Calculate  the  efficiency  of  the  rotary  A  when  it  delivers  300 
kw.  at  625  volts  and  the  excitation  is  greater  than  normal  and  is 
adjusted  so  that  the  line  current  on  the  alternating-current  side 
is  520  amperes.  Assume  that  the  running  temperature  is  70°  C. 

1  In  calculating  the  armature  reaction  do  not  neglect  the  distribution  of 
the  winding,  and  assume  that  the  constant  usually  given  as  0 .  707  is  0 . 60. 
This  makes  an  approximate  correction  for  the  effect  of  the  ratio  of  pole  arc 
to  pole  pitch.  As  a  first  approximation  it  may  be  assumed  that  the  effi- 
ciency is  95  per  cent. 

8 


104  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 


DATA  ON  ROTARY  CONVERTERS 


A 

B 

C 

Kilowatt  

300 

1000 

1000 

Voltage  (D.C.)  

600 

600 

600 

Current  (D.C.) 

500 

1667 

1AA7 

Phases  

3 

6 

Q 

Frequency 

25 

25 

60 

Poles  

4 

8 

12 

Armature  slots  

96 

168 

180 

Inductors  per  slot  
Turns  per  pole: 
Shunt.  

6 
2340 

6 
1501 

6 
864 

Series      

9 

2 

2 

Resistance  at  25°  C.  : 
Armature  between  D.C.  brushes  
Shunt  field2  (alone)  

0.030 
120.5 

0.00688 
76  6 

0.00589 
39  65 

Series  field  
Friction  and  windage  (kw.)  

0.00248 
3.2 

0.00042 
6.7 

0.0006095 
9.1 

D.  C.  SATURATION  CURVES  AND  CORE  LOSSES 


Ordinates 

Abscissae 

D.C. 

voltage 

Shunt 

field  currents 

Core 

losses 

A 

B 

C 

A 

I 

B       \       C 

550 
600 
650 
700 

2 

3 
4 
5 

.70 

.25 
.00 
.00 

4 
5 
6 

7 

.90 

.60 
.36 
.44 

7 
9 
11 
14 

9 
3 
2 
0 

1.96 
2.52 
3.25 
4.23 

4 
5 
6 

7 

.68 
.50 
.40 
.70 

12. 

14. 
17. 
21. 

4 

7 
7 
6 

60. l  Calculate  the  efficiency  of  the  rotary  B  when  it  delivers 
1000  kw.  at  650  volts  and  the  shunt  field  rheostat  is  cut  out. 
Assume  that  the  running  temperature  is  70°  C.  What  are  the 
line  current  and  power  factor  on  the  alternating-current  side? 

61.  *  Calculate  the  efficiency  of  the  rotary  C  when  it  delivers 
1000  kw.  at  650  volts  and  the  excitation  is  greater  than  normal 
and  is  adjusted  so  that  the  power  factor  is  0.95.  Assume  that 
the  running  temperature  is  70°  C. 

1  In  calculating  the  armature  reaction  do  not  neglect  the  distribution  of 
the  winding,  and  assume  that  the  constant  usually  given  as  0.707  is  0.60. 
This  makes  an  approximate  correction  for  the  effect  of  the  ratio  of  pole  arc 
to  pole  pitch. 

2  The  shunt  field  has  a  regulating  rheostat  in  series  with  it,  the  loss  in 
whichjshould  be  included  in  calculating  the  efficiency. 


CHAPTER  VI 
POLYPHASE  CIRCUITS 

1.  Three  equal  impedance  units,  each  of  which  has  an  equiva- 
lent resistance  of  2.0  ohms  and  a  reactance  of  1.25  ohms  are 
connected  in  delta  across  a  three-phase  220-volt  circuit.     What 
current  does  each  unit  take?     What  is  the  line  current?  What  is 
the  total  power  supplied? 

2.  The  three  impedance  units  described  in  problem  1  are  con- 
nected in  Y  across  a  three-phase,  220-volt  circuit.     What  cur- 
rent does  each  unit  take?     What  is  the  total  power  supplied? 

3.  Six  equal  impedance  units  each  of  which  has  an  equivalent 
resistance  of  2.5  ohms  and  a  reactance  of  1.5  ohms  are  connected 
across  a  three-phase,  220-volt  circuit — three  in  delta  and  three  in  Y. 
What  is  the  line  current?     What  is  the  total  power  supplied? 

4.  Three  equal  impedance  units  each  of  which  has  an  equivalent 
resistance  of  2  ohms  and  a  condensive  reactance  of  1  ohm  are 
connected  in  delta  across  a  three-phase,  220-volt  circuit.     At  the 
same  point  three  other  equal  impedance  units,  each  of  which  has 
an  equivalent  resistance  of  1.5  ohms  and  an  inductive  reactance 
of  1  ohm,  are  connected  in  Y  across  the  circuit.     What  is  the  line 
current?     What  is  the  total  power  supplied? 

5.  Three  equal  resistances  are  connected  in  delta  across  a  three- 
phase  circuit.     What  should  be  the  relative  value  of  three  other 
equal  resistances  which  will  take  the  same  power  when  connected 
in  Y  across  the  circuit? 

6.  Three  equal  impedance  units,  each  of  which  has  an  equiva- 
lent resistance  of  2.0  ohms  and  a  reactance  of  1.0  ohm  are  con- 
nected in  delta  across  a  three-phase,  220-volt  circuit.     Three 
other  equal  impedance  units  are  connected  in  Y  across  the  same 
circuit.     What  should  be  their  equivalent  resistance  and  react- 
ance in  order  that  they  will  take  the  same  line  current  and  the 
same  total  power? 

7.  Three  equal  impedance  units  each  of  which  has  an  equiva- 
lent resistance  of  2.0  ohms  and  an  inductive  reactance  of  1.0 
ohm  are  connected  in  delta  at  the  end  of  a  transmission  line,  each 
conductor  of  which  has  a  resistance  of  0.2  ohm  and  an  inductive 

105 


106  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

reactance  of  0.3  ohm.  If  the  line  voltages  at  the  generating 
station  are  each  2200  volts  what  is  the  line  current?  (b)  What 
is  the  voltage  at  the  load?  (c)  What  is  the  efficiency  of  trans- 
mission? 

8.  In  problem  7  if  the  reactance  of  the  impedance  units  had  been 
condensive  instead  of  inductive  what  would  have  been  (a)  the 
line  current,  (b)  the  voltage  at  the  load,  and  (c)  the  efficiency  of 
transmission? 

9.  Two  equal  resistances  of  100  ohms  each  are  connected  in 
series  across  two  mains  of  a  three-phase  220-volt  circuit  and 
from  their  junction  a  resistance  of  50  ohms  is  connected  to  the 
neutral  conductor  of  the  system.     The  line  voltages  are  balanced 
and  the  voltages  from  the  lines  to  the  neutral  conductor  are 
equal,     (a)  What  are  the  line  currents?     (b)  What  is  the  neutral 
current?     (c)  What  is  the  total  power  absorbed? 

10.  Two  equal  impedances,  each  of  which  has  an  equivalent 
resistance  of  2.0  ohms  and  an  inductive  reactance  of  1.0  ohm 
are  connected  in  series  across  two  mains  of  a  three-phase,  220-volt 
circuit,  and  from  their  junction  another  unit  which  has  a  resist- 
ance of  1.0  ohm  and  a  condensive  reactance  of  1.0  ohm  is  connected 
to  the  neutral  conductor  of  the  system.     The  line  voltages  are 
balanced  and  the  voltages  from  the  lines  to  the  neutral  are  equal, 
(a)  What  are  the  line  currents?     (b)  What  is  the  neutral  current? 
(c)  What  is  the  total  power  absorbed? 

11.  Three  non-inductive  resistances  of  5,  10  and  15  ohms  are 
connected  in  delta  across  the  lines  of  a  three-phase,  220-volt 
circuit,     (a)  What  is  the  total  power  absorbed?     (b)  What  are 
the  line  currents? 

12.  Three   impedance   units   which   are   represented   by   the 
expressions,  Zi  =  5+./5,  z2  =  5+j!Q,  z3  =  5  —  zjlQ,  are  connected  in 
delta  across  the  lines  1-2,  2-3,  3-1  respectively  of  a  three-phase, 
220-volt  circuit.     If  Viz  leads  VM  by  120  degrees  (a)  what  is 
the  total  power  absorbed?     (b)  What  are  the  line  currents? 

If  Viz  lags  F23  by  120  degrees,  (c)  what  is  the  total  power 
absorbed?     (d)  What  are  the  line  currents? 

13.  Three  non-inductive  resistances  of  5,  10  and  15  ohms  are 
connected  in  Y  across  the  lines  of  a  three-phase,  220-volt  circuit. 
(a)   What  is  the  total  power  absorbed?     (b)  What  are  the  line 
currents? 

14.  Three  impedance  units  which  are  represented  by  the  ex- 
pressions, zi  —  5+J5,  22  =  5+jlO,  23  =  5  —  j'10,  are  connected  in  Y 


POLYPHASE  CIRCUITS  107 

from  the  mains  1,  2  and  3  respectively  of  a  three-phase,  220-volt 
circuit  to  a  common  point.  If  Vi2  leads  Vzs  by  120  degrees,  (a) 
what  are  the  line  currents?  (b)  What  is  the  total  power  absorbed? 
(c)  What  is  the  voltage  between  the  "common  point"  and  the 
true  neutral  of  the  system? 

15.  In  problem  14  if  F23  leads  Viz  by  120  degrees  a  what  are 
the  line  currents?     (b)  What  is  the  total  power  absorbed?     (c) 
What  is  the  voltage  between  the  " common  point"  and  the  true 
neutral  of  the  circuit? 

16.  Three  impedance  units  which  are  represented  by  the  ex- 
pressions Zi=10-fjO,  22  =  0+jlO,  03  =  0— jW,  are  connected  in 
Y  from  the  mains  1,  2  and  3  respectively  of  a  three-phase, 
220-volt  circuit  to  a  common  point.     If  Viz  leads  Vzs  by  120  de- 
grees (a)  what  are  the  line  currents?     (b)  What  is  the  total  power 
absorbed?     (c)  What  is  the  voltage  between  the  "  common  point" 
and  the  true  neutral  of  the  system? 

17.  In  problem  16  if  Vzz  leads  Viz  by  120  degrees  (a)  what  are 
the  line  currents?     (b)What  is  the  total  power  absorbed?     (c) 
What  is  the  voltage  between  the  "common  point"  and  the  true 
neutral  of  the  system? 

18.  Three  voltmeters  are  connected  in  Y  from  the  mains  of  a 
three-phase,  220-volt  circuit  to  a  common  point.     The  resistances 
of  the  voltmeters  are  Ri  ohms  and  Rz  ohms  and  R  3  ohms  respec- 
tively.    What  does  each  instrument  indicate? 

19.  Three  unequal   lamp   loads  are   connected  between  the 
mains  and  neutral  conductor  of  a  three-phase  transmission  line. 
The  mains  and  neutral  conductor  each  have  a  resistance  of  0.1 
ohm   and    negligible  reactance.     The  resistances   of  the  lamp 
loads  are  1.0  ohm,  1.5  ohms   and   2.0  ohms  respectively.     At 
the  generating  station  the  line  voltages  are  each  220  volts  and  the 
voltages  between  the  mains  and  neutral  conductors  are  equal, 
(a)  What  are  the  line  and  neutral  currents?     (b)  What  is  the 
voltage  across  each  lamp  load?     (c)  What  is  the  efficiency  of 
transmission? 

20.  Three  unequal  lamp  loads  are  connected  in  delta  at  the 
end  of  a  three-phase  transmission  line  which  has  a  resistance  of 
0.1  ohm  per  conductor  and  negligible  reactance.     The  resistances 
of  the  lamp  loads  are  1.0  ohm,  1.5  ohms  and  2.0  ohms  respectively. 
If  the  line  voltages  at  the  generating  station  are  each  220  volts, 
(a)  what  are  the  line  currents?     (b)  What  are  the  line  voltages  at 
the  load?     (c)  What  is  the  efficiency  of  transmission? 


108  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

21.  Three  impedance  units  whose  values  are  represented  by 
the  expressions,   2i  =  2.0+./   1.0,    22  =  1.5+.?   0.5,  and  z3  =  2.5  + 
j  1.0,  are  connected  in  delta  at  the  end  of  a  three-phase  transmis- 
sion line  each  conductor  of  which  has  a  resistance  of  0.1  ohm  and 
a  reactance  of  0.15  ohm.     If  the  line  voltages  at  the  generating 
station  are  each  11,000  volts,  (a)  what  are  the  line  currents? 
(b)  What  are  the  line  voltages  at  the  load?     (c)  What  is  the  effi- 
ciency of  transmission? 

22.  The  power  supplied  to  a  three-phase  induction  motor  is 
measured  by  the  two- wattmeter  method.     One  wattmeter  indi- 
cates 5770  watts  and  the  other  2930  watts.     What  is  the  power 
supplied?     At  what  power  factor  is  the  motor  operating? 

23.  The  only  instrument  available  for  measuring  the  power 
taken  by  a  three-phase,  230-volt  induction  motor  is  a  wattmeter 
of  suitable   range.     Measurements  are   made   as   follows:  The 
current  coil  of  the  wattmeter  is  inserted  in  main  1  and  the  poten- 
tial coil,  first  between  mains  1  and  2  and  then  between  mains 
1  and  3.     If  the  two  wattmeter  readings  thus  obtained  are  5760 
and  3380  watts  respectively  what  is  the  power  supplied  to  the 
motor?     If  the  line  voltage  is  230  volts  what  is  the  line  current? 
At  what  power  factor  is  this  motor  operating? 

24.  The  power  taken  by  a  balanced  three-phase  load  is  meas- 
ured by  two  wattmeters.     The  current  coils  of  the  wattmeters  are 
connected  to  current  transformers  which  are  in  lines  1  and  2 
respectively.     The  potential   coils  are   connected   to  potential 
transformers  which  are  across  lines  2  and  3  and  lines  1  and  3 
respectively.     The  line  voltages  are  each  230  volts  and  the  line 
currents  are  each  150  amperes.     The  wattmeters  each  indicate  19.6 
kw.     What  is  the  power  supplied?     What  is  the  power  factor? 

25.  The  power  taken  by  an  unbalanced  three-phase  load  is 
measured  by  two  wattmeters.     The  current  coils  of  the  watt- 
meters are  connected  to  current  transformers  which  are  in  lines 
1  and  2  respectively,  and  the  potential  coils  are  connected  to 
potential  transformers  which  are  across  lines  2  and  3  and  lines  1 
and  3  respectively.     The  line  voltages  are  each  230  volts.     The 
currents  in  lines  1  and  2  are  150  amperes  and  200  amperes  re- 
spectively.    The  first  wattmeter  indicates  21. 2  kilowatts  and  the 
second  indicates  18.1  kilowatts.     What  is  the  power  supplied  to 
the  load? 

26.  A    3-phase,    500-volt,     Y-connected    alternating-current 
generator  with  equal  line  voltages  and  a  grounded  neutral  supplies 


POLYPHASE  CIRCUITS  109 

energy  to  an  unbalanced  Y-connected  load,  the  neutral  of  which  is 
not  grounded.  The  line  currents  are,  /i  =  141.4  amperes,  72  =  100 
amperes  and  73  =  100  amperes.  One  wattmeter  is  used  and  it  is 
connected  with  its  current  coil  in  line  1  and  its  potential  coil 
across  lines  1  and  2.  If  this  wattmeter  indicates  70.7  kw.  what 
is  the  total  power  supplied  to  the  load?  If  the  power  factors 
of  each  of  the  three  phases  are  equal  what  is  the  voltage  between 
the  neutral  of  the  load  and  the  neutral  of  the  generator?  What 
is  the  power  factor? 

27.  An  unbalanced  lamp  load,  consisting  of  115-volt  lamps,  is 
connected  in  Y  across  the  lines  of  a  balanced  3-phase,  200-volt 
circuit.     The  line  currents  are  70.7  amperes,  50  amperes  and  50 
amperes  respectively.     What  is  the  power  supplied  to  this  load? 
What  is  the  voltage  across  each  phase  of  the  load? 

28.  In  problem   27  i£  the    resistances  of  the  lamp  load  are 
assumed  to  be  constant,  what  will  be  the  currents  in  the  lines  and 
neutral  when  the  neutral  point  of  the  load  is  connected  to  the 
neutral  conductor  of  the  circuit?     The  voltages  between  the 
neutral  conductor  and  the  lines  are  equal. 

29.  An  unbalanced  lamp  load  is  connected  in  delta  across  the 
lines  of  a  balanced  3-phase,  230-volt  circuit.     The  resistances 
of  these  loads  between  lines  1  and  2,  2  and  3,  and  3  and  1  are  10 
ohms,  8  ohms  and  6  ohms  respectively,     (a)  What  are  the  line 
currents?     (b)  If  the  power  is  measured  by  two  wattmeters  which 
have  their  current  coils  in  lines  1  and  2  what  will  each  instrument 
indicate? 

30.  An  unbalanced  lamp  load,  consisting  of  115-volt  lamps,  is 
connected  between  the  lines  and  neutral  conductor  of  a  balanced 
3-phase,  200-volt  circuit.     The  resistances  of  the  loads  between 
lines  1,  2,  and  3  and  the  neutral  conductor  are  6,  8,  and  10  ohms 
respectively.     The  line  voltages  are  equal  and  the  voltages  from 
the  lines  to  the  neutral  conductor  are  also  equal.     What  would 
two  wattmeters  indicate  which  have  their  current  coils  in  lines 
1  and  2  and  their  potential  coils  across  lines  1  and  3  and  lines  2  and 
3  respectively?     What  is  the  total  power  supplied?     What  is  the 
current  in  the  neutral  conductor? 

31.  Three  single-phase  transformers  each  of  which  has  a  ratio 
of  transformation  of  2.5  to  1  are  connected  in  delta  on  the  high- 
tension  side.     The  low-tension  windings  are  not  connected  but 
supply  three  separate  single-phase  loads.     The  first  of  these  loads 
is  90  kilowatts  at  unit  power  factor,  the  second  is  60  kilowatts 


110  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

at  0.7  power  factor,  and  the  third  is  30  kilowatts  at  unit  power 
factor.  The  high-tension  line  voltages  are  600  volts.  Neglect 
the  losses  in  the  transformers.  What  are  the  high-tension  line 
currents? 

32.  Three  unequal    single-phase  motor  loads    are  connected 
across  the  lines  of  a  balanced  3-phase,  230-volt  circuit.     The  first 
takes  106  amperes  at  0.78  power  factor,  the  second  takes  142 
amperes  at  0.82  power  factor,  and  the  third  takes  28.4  kilowatts 
at  0.77  power  factor.     What  are  the  line  currents? 

33.  Three  unequal    single-phase    motor  loads    are  connected 
between  the  lines  and  neutral  conductor  of  a  balanced  3-phase, 
350-volt  circuit.     The  voltages  from  the  lines  to  the  neutral  are 
each  202  volts.     The  first  load  takes  20  kilowatts  at  0.82  power 
factor,  the  second  takes  28  kilowatts  at  0.75  power  factor,  and  the 
third  takes  36  kilowatts  at  0.79  power  factor.     What  is  the 
current  in  the  neutral  conductor? 

34.  From   the    terminals   of   a   3-phase,    550-volt,    60-cycle 
alternating-current  generator  runs  an  artificial  transmission  line 
which  has  a  resistance  of  0.1  ohm  and  an  inductance  of  1.0  milli- 
henry per  conductor.     At  the  end  of  this  line  is  a  balanced  load  of 
three  reactors  connected  in  Y.     The  equivalent  resistance  and 
reactance  of  these  reactors  should  be  assumed  to  be  constant. 
The  power  output  of  the  generator  is  measured  by  the  two-watt- 
meter method  with  the  current  coils  of  the  instruments  in  lines  1 
and  2.     The  first  wattmeter  indicates  40  kw.  and  the  second, 
100  kw. 

If  line  3  is  opened  at  the  load  what  power  will  the  instruments 
indicate?     The  terminal  voltage  of  the  generator  is  constant. 

35.  In  problem  34  if  line  3  is  opened  between  the  terminal 
of  the  generator  and  the  potential  coils  of  the  wattmeter,  what 
power  will  the  instruments  indicate? 

36.  From  the  terminals  of  a  3-phase,  500-volt,  25-cycle  alter- 
nating-cnrrent  generator  runs  an  artificial  transmission  line  which 
has  a  resistance  of  0.3  ohm  and  an  inductance  of  3  millihenrys 
per  conductor.     At  the  end  of  this  line  is  a  balanced  load  of  three 
reactors   connected   in   delta.     The   equivalent   resistance    and 
reactance  of  these  reactors  should  be  assumed  to  be  constant. 
The  power  output  of  the  generator  is  measured  by  the  two- 
wattmeter  method  with  the  current  coils  of  the  instruments  in 
lines  1  and  2.     The  first  wattmeter  indicates  50  kw.  and  the 
second,  25  kw. 


POLYPHASE  CIRCUITS  111 

If  line  3  is  opened  what  power  will  the  instruments  indicate? 
The  terminal  voltage  of  the  generator  is  constant. 

37.  In  problem  36  if  line  3  is  opened  between  the  terminal  of 
the  generator  and  the  potential  coils  of  the  wattmeters  what 
power  will  each  instrument  indicate?     The  terminal  voltage  of 
the  generator  is  constant. 

38.  Two  alternating-current  generators,  operating  in  parallel, 
deliver  power  to  a  balanced,  3-phase  load.     The  output  of  each 
generator   is   measured   by   the   two-wattmeter  method.     The 
terminal  voltage  is  2210  volts.     The  wattmeter  readings  are: 

First  generator  Second  generator 

TFi  =  196kw.  TFi  =  172kw. 

TF2  =  312kw.  TF2  =  88kw. 

The  similarly  numbered  instruments  are  connected  in  the  same 
lines.  What  is  the  total  power  supplied?  What  is  the  power 
factor  of  the  load? 

39.  Two  alternating-current  generators,  operating  in  parallel, 
deliver  power  to  a  balanced  3-phase  load.     The  output  of  each 
generator  is  measured  by  the  two-wattmeter  method.     The  ter- 
minal voltage  is  6650  volts.     The  wattmeter  readings  are: 

First  generator  Second  generator 

W i  =  412  kw.  Wi  =  183  kw. 

TF2  =  626  kw.  W 2  =  457  kw. 

The  similarly  numbered  instruments  are  connected  in  the  same 
lines.  What  is  the  total  power  supplied?  What  is  the  power 
factor  of  the  load? 

40.  It  is  desired  to  transform  200  kw.  from  2-phase  to  3-phase 
by  Scott-connected  transformers.     The  two-phase  line  voltage 
is  2200  volts  and  the  three-phase  line  voltage  is  230  volts.     What 
should  be  the  current  and  voltage  rating,  and  the  ratio  of  trans- 
formation of  each  transformer? 

41.  It  is  desired  to  transform  100  kw.  from  3-phase  at  6600 
volts  to  2-phase  at  110  volts  by  the  means  of  Scott-connected 
transformers.     What  should  be  the  current  and  voltage  rating, 
and  the  ratio  of  transformation  of  each  transformer? 

42.  Three  single-phase  transformers  are  connected  in  delta  on 
both  the  primary  and  secondary  sides.     With  a  primary  voltage 
on  open  circuit  of  22,000  volts  the  secondary  voltage  is  440  volts. 
The  short-circuit  characteristic  data  of  each  transformer  are: 


112  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

V  =  1020  volts,  /=  1.136  amperes,  P  =  351  watts.  If  the  second- 
ary terminal  voltage  is  440  volts  when  there  is  a  single-phase 
load  taking  40  kw.  at  0.8  power  factor  connected  across  one  phase 
of  the  secondary,  what  will  be  the  percentage  rise  in  this  voltage 
when  the  load  is  thrown  off? 

43.  (a)  On  the  basis  of  the  same  heating  loss  compare  the 
full-load  kilowatt  output  at  unit  power  factor  of  two  single-phase 
transformers  connected  in  open  delta  with    their    name-plate 
rating,     (b)  On  the  basis  -of  the  same  heating  loss  compare  the 
full-load  kilowatt  output  at  0.87  power  factor — both  lagging  and 
leading — of  two  single-phase  transformers  connected  in  open  delta 
with  87  per  cent,  of  their  name-plate  rating.     In  each  case  the  load 
is  balanced. 

44.  (a)  Three  single-phase  transformers  with  both  primaries 
and  secondaries  connected  in  delta  supply  a  balanced  load  of  100 
kw.  at  unit  power  factor.     If  it  is  necessary  to  remove  one  of  these 
transformers  from  the  line  by  what  per  cent,  will  the  copper 
loss  in  each  of  the  other  two  transformers  be  increased?     (b) 
If  the  load  had  been  a  balanced  one  taking  87  kw.  at  0.87  power 
factor — both  leading  and  lagging  -what  would  have  been  the 
per   cent,    increase   in  the    copper   loss  in  each   of  the   other 
transformers? 

45.  (a)  On  the  basis  of  the  same  copper  loss  in  each  trans- 
former compare  the  full-load  kilowatt  output  at  unit  power 
factor  of  two  equal  single-phase  transformers  connected  in  T 
with  their  name-plate  rating,     (b)  On  the  basis  of  the  same  copper 
loss  in  each  transformer  compare  the  full-load  kilowatt  output 
at  0.87   power  factor  of  two  equal  single-phase  transformers 
connected  in  T  with  87  per  cent,  of  their  name-plate  rating. 

46.  On  the  basis  of  the  same  copper  loss  in  each  transformer 
compare  the  full-load  kilowatt  output  of  two  equal  single-phase 
transformers  connected  in  open  delta  with  the  full-load  output  ol 
the  same  transformers  connected  in  T  (a)  at  unit  power  factor, 
(b)  at  0.87  power  factor — both  lagging  and  leading. 

47.  Compare  the  regulation  of  three   lOO-kv.-a.  transformers 
whose  primaries  and  secondaries  are    connected  in  delta  with 
that  of  two  100-kv-a.  transformers  whose  primaries  and  second- 
aries are  connected  in  open  delta  when  a  single-phase  load  of 
100  kw.  at  0.8  power  factor  is  delivered  on  the  secondary  side. 
The  secondary  or  low-tension  voltage  under  load  conditions  is  2200 
volts  and  the  ratio  of  transformation  is  5  to  1.     The  short-circuit 


POLYPHASE  CIRCUITS  113 

of  each  transformer  are:  F  =  310  volts,  7  =  9.1  amperes  (full-load) 
P  =  1000  watts.  In  the  case  of  the  open  delta  the  load  is  connected 
across  one  transformer. 

48.  In  problem  47  compare  the  regulation  in  the  two  cases 
when  a  single-phase  load  of  150  kw.  at  0.8  power  factor  is  sup- 
plied, which  in  the  case  of  the  open  delta  is  connected  across  the 
terminals  of  the  two  transformers,  i.e.,  across  the  open  side. 

49.  Three  transformers  whose  primaries  are  connected  in  Y 
and  whose  secondaries  are  connected  in  delta  are  in  parallel  on  the 
primary  side  with  three  others  whose  primaries  and  secondaries 
are  both  connected  in  Y.     If  one  secondary  terminal  of  the  first 
set  is  connected  to  one  corresponding  terminal  of  the  second  set 
what  are  the  greatest  and  least  voltages  that  can  exist  between 
the  other  secondary  terminals  of  the  two  sets?     The  line  voltages 
on  the  secondary  sides  are  1 100  volts  for  the  two  sets  of  transformers. 

50.  Three  transformers  whose  primaries  and  secondaries  are 
both  connected  in  delta  are  in  parallel  on  the  primary  side  with 
three  others  whose  primaries  are  connected  in  delta  and  whose 
secondaries  are  connected  in  Y.     If  one  secondary  terminal  of 
the  first  set  is  connected  to  a  corresponding  terminal  of  the 
second  set,  what  are  the  greatest  and  least  voltages  that  can  exist 
between  the  other  secondary  terminals  of  the  two  sets.     The  line 
voltages  are  1100  volts  for  the  two  sets  of  transformers. 

51.  Three  transformers  whose  primaries  and  secondaries  are 
both  connected  in  delta  are  in  parallel  on  the  primary  side  with 
three  others  whose  primaries  are  connected  in  Y  and  whose  second- 
aries are  connected  in  delta.     If  one  secondary  terminal  of  the  first 
set  is  connected  to  a  corresponding  terminal  of  the  second  set 
what  are  the  greatest  and  least  voltages  that  can  exist  between 
the  other  secondary  terminals  of  the  two  sets.     The  line  voltages 
are  1100  volts  for  the  two  sets  of  transformers. 

52.  Three  auto-transformers  are  connected  as  shown  in  Fig.  3 
to  receive  power  from  a  3-phase,  11,000-volt  circuit.     The  ratio 
of  transformation  for  each  transformer  from  high  tension  to  low 
tension  is  2  to  1.     What  is  the  secondary  line  voltage,  and  what  is 
the  phase  relation  of  the  corresponding  primary  and  secondary 
line  voltages  on  open  circuit? 

53.  Three  auto-transformers  are  connected  as  shown  in  Fig.  4 
to  receive  power  from  a  3-phase,  11,000-volt  circuit.     The  ratio 
of  transformation  f ron  high  tension  to  low  tension  is  2  to  1 .     What 
is  the  secondary  line  voltage,  and  what  is  the  phase  relation  of 


114  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 


the  corresponding  primary  and  secondary  line  voltages  on  open 
circuit? 

54.  Three  auto-transformers  are  connected  as  shown  in  Fig.  5 
to  receive  power  from  a  3-phase,  11,000-volt  circuit.  The  ratio 
of  transformation  for  each  transformer  from  high  tension  to  low 
tension  is  2  to  1.  What  is  the  secondary  line  voltage,  and  what 


L.T. 


L.T. 


H.T. 


FIG.  4. 


is  the  phase  relation  of  the  corresponding  primary  and  secondary 
line  voltages  on  open  circuit? 

55.  A  1500-kw.,  5500-volt,  3-phase  generator  delivers  power 
to  a  transmission  line  through  three  single-phase  transformers 
which  have  their  low-tension  windings  connected  in  delta  and  their 
high-tension  windings  connected  in  Y.  The  following  data  on  this 
generator  and  the  transformers  are  given: 

GENERATOR 


Field  current 

Open-circuit  terminal 
voltage 

Short-circuit  armature 
current 

150 
200 
250 

5100 
5900 
6500 

300 
400 

300 

6800 

350 

7100 

The  core  loss  at  the  rated  voltage  is  20.2  kw.  and  the  friction 
and  windage  is  8.4  kw.  Both  of  these  losses  may  be  assumed 
constant.  The  effective  resistance  of  the  armature  is  0.362  ohm 
per  phase.  The  resistance  of  the  field  winding  is  0.376  ohm. 
The  armature  windings  are  connected  in  Y. 

TRANSFORMER 


Voltage 


Short  circuit 


-ivv.-a, 

High  tension 

Low  tension 

Amperes1 

Volts 

Watts 

500 

12,700 

5500 

39.4 

332 

4680 

1  Full-load  current. 


POLYPHASE  CIRCUITS 


115 


The  core  loss  at  the  rated  voltage  is  3330  watts  and  may  be 
assumed  constant. 

(a)  When  the  power  delivered  to  the  transmission  is  1280  kw. 
at  a  power  factor  of  0.83  and  a  line  voltage  of  22,300  volts,  what 
is  the  combined  efficiency  of  the  generator  and  transformers? 
Calculate  the  generator  field  current  by  the  magnetomotive-force 
method. 

(b)  If  this  load  is  removed  from  the  line  and  the  field  excita- 
tion of  the  generator  is  unchanged,  what  is  the  high-tension  line 
voltage? 

56.  A  1640-kw.,  13,500-volt,  3-phase  generator  delivers 
power  to  a  transmission  line  through  three  single-phase  trans- 
formers which  are  connected  in  Y  on  both  the  high-  and  low- 
tension  sides.  The  following  data  are  given  on  this  generator 
and  transformers: 

GENERATOR 


Field  current 

Open-circuit  terminal 
voltage 

Short-circuit  armature 
current 

50 
100 
150 
200 

7,500 
10,100 
14,700 
15  800 

75 
155 

227 

250 

16.700 

The  core  loss  at  the  rated  voltage  is  21.3  kw.,  and  the  friction 
and  windage  is  8.8  kw.  Both  of  the  losses  may  be  assumed  con- 
stant. The  effective  resistance  of  the  armature  is  1.52  ohms  per 
phase.  The  resistance  of  the  field  circuit  is  0.392  ohms.  The 
armature  windings  are  connected  in  Y. 

TRANSFORMERS 


Kv.-a 

Voltage 

Short  circuit 

High  tension 

Low  tension 

Amperes1 

Volts 

Watts 

500 

38,000 

7800 

13.15 

1190 

3375 

The  core  loss  at  the  rated  voltage  is  2960  watts,  and  may  be 
assumed  constant. 

(a)  When  the  output  of  the  generator  is  1460  kw.,  at  0.92 
power  factor  and  a  terminal  voltage  of  13,600  volts  what  is  the 
combined  efficiency  of  the  generator  and  transformers?  Cal- 
culate field  current  of  the  generator  by  the  magnetomotive-force 
method. 

1  Full-load  current. 


116  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

(b)  If  this  load  is  removed  from  the  line  and  the  field  excita- 
tion of  the  generator  is  unchanged  what  is  the  high-tension  line 
voltage? 

57.  A  1000-kw.,  2400-volt,  3-phase  generator  delivers  power 
to  a  transmission  line  through  three  single-phase  transformers 
which  are  connected  in  delta  on  the  low-tension  and  in  Y  on  the 
high-tension  side.     Each  transformer  has  a  ratio  of  transforma- 
tion of  5.29  to  1.     The  resistances  of  the  high-  and  low-tension 
windings  are  2.02  ohms  and  0.072  ohm  respectively.     The  effective 
resistance  of  the  generator  is  0.067  ohm  per  phase.     The  armature 
windings  are  connected  in  Y.     With  the  high-tension  windings 
of  the  transformers  short  circuited  and  with  a  field  excitation 
of  100  amperes  for  the  generator  the  armature  current  is  458 
amperes.     With  this  same  excitation  the  open-circuit  terminal 
voltage  of  the  generator  is  2220  volts.     The  rotation  losses  in  the 
generator  are  31.2  kw.  at  normal  voltage   and  the  core  losses 
in  each  transformer  are  1.8,  kw. 

(a)  What  is  the  combined  efficiency  of  the  generator  armature 
and  the  transformers  when  a  balanced  load  of  954  kw.  at  0.91 
power  factor  is  delivered  on  the  high-tension  side  of  the  trans- 
formers at  a  line  voltage  of  22,400  volts? 

(b)  What  would  be  the  high-tension  line  voltage  if  this  load 
were  removed  and  the  excitation  of  the  generator  unchanged? 

58.  A  760-kw.,  2200-volt,  3-phase  generator  delivers  power  to 
a  transmission  line  through  three  single-phase  transformers  which 
have  both  their  high  and  low-tension  windings  connected  in  Y. 
With  the  high-tension  windings  of  the   transformers   short-cir- 
cuited the  output  of  the  generator  is  37.8  kilowatts,  the  armature 
current  is  450  amperes  when  the  terminal  voltage  is  133  volts. 
With  the  transformers  on  open  circuit  and  with  the  same  field 
excitation,  the  terminal  voltage  of  the  generator  is  1780  volts. 
The  effective  resistance  of  the  armature  is  0.172  ohm  per  phase. 
The  armature  windings  are  connected  in  Y.     The  rotation  losses 
of  the  generator  are  17.2  kw.  at  normal  voltage  and  the  core  loss 
in  each  transformer  is  1670  watts.     The  transformers  have  a 
ratio  of  transformation  of  10  to  1. 

(a)  What  is  the  combined  efficiency  of  the  generator  armature 
and  the  transformers  when  a  balanced  load  of  680  kw.  at  0.9 
power  factor  is  delivered  to  the  transmission  line  at  22,400  volts? 

(b)  What  would  be  the  high-tension  line  voltage  if  this  load 
were  removed  and  the  excitation  of  the  generator  unchanged? 


POLYPHASE  CIRCUITS  117 

59.  A  1000-kw.,  11,000-volt,  3-phase  generator  delivers  power 
to  a  transmission  line  at  the  end  of  which  is  a  1200-h.p.  induction 
motor.     With  the  induction  motor  running  at  no  load  the  exci- 
tation of  the  generator  is  adjusted  so  that  the  terminal  voltage, 
the  line  current  and  the  total  power  measured  at  the  motor  are 
10,600  volts,  20.2  amperes  and  20.4  kilowatts  respectively.     At 
the  same  time  the  terminal  voltage  and  the  total  power  measured 
at  the  generator  are  10,980  volts  and  29.8  kw.     The  field  excita- 
tion of  the  generator  is  adjusted  so  that,  when  the  motor  is  deliver- 
ing full  load,  its  terminal  voltage  is  11,000  volts.     The  efficiency 
and  power  factor  of  the  motor  at  full  load  are  0.921  and  0.906 
respectively. 

What  is  the  terminal  voltage  of  the  generator  when  the  motor 
delivers  full  load?  What  is  the  efficiency  of  transmission  at  this 
time? 

60.  A  1000-kw.,  13,800-volt,  3-phase  generator  delivers  power 
directly  to  a  transmission  line  at  the  end  of  which  is  an  induction 
motor  load.     The  resistance  and  reactance  of  the  transmission 
line  are  16.8  ohms  and  17. 2  ohms  per  conductor.     The  generator 
has  an  effective  armature  reactance  of  2.18  ohms  per  phase.     The 
field  current  is  supplied  at  120  volts. 


Field 
current 

Open-circuit  terminal 
voltage1 

Terminal  voltage  with  an  armature 
current  of  42  amp.  at  zero 
power  factor 

50 

8,800 

80 

13,000 

110 
140 
180 

15,600 
17,250 
18,900 

10,750 
13,250 
15,600 

On  open  circuit  the  rotational  losses  are  16.6  kw.  and  25.4  kw. 
when  the  terminal  voltages  are  13,000  and  15,600  volts  respec- 
tively. 

What  should  be  the  field  excitation  of  the  generator  so 
that  the  line  voltage  at  the  motor  load  will  be  13,200  volts  when 
the  motors  take  926  kw.  at  0.91  power  factor?  What  is  the  effi- 
ciency of  the  generator  and  the  line? 

61.  At  the  end  of  a  3-phase  transmission  line  is  a  motor  load 
requiring  3000  kw.  The  line  voltage  at  the  load  should  be  32,000 
volts  and  the  power  factor  of  the  load  is  0.90.  What  should  be 

1  The  generator  is  Y-connected. 


118  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

the  resistance  and  reactance  of  the  line  per  conductor  so  that 
the  efficiency  of  transmission  will  be  90  per  cent,  and  the  voltage 
regulation,  12  per  cent? 

62.  A   1640-kv.-a.,    13,500-volt,    3-phase    generator    delivers 
power  directly  to  a  transmission  line  which  has  a  resistance  of 
30.2  ohms  and  a  reactance  of  24  ohms  per  conductor.     With  the 
far  end  of  the  line  short-circuited  and  with  a  field  excitation  of  150 
amperes  the  line  current  is  138  amperes.     With  the  same  field 
excitation  the  open-circuit  terminal  voltage  of  the  generator  is 
14,700  volts.     The  effective  resistance  of  the  armature  is  1.52 
ohms  per  phase.     The  generator  is  Y-connected. 

What  is  the  combined  electrical  efficiency  of  the  armature  of 
the  generator  and  the  transmission  line  for  a  load  of  1500  kw. 
at  0.90  power  factor  if  the  line  voltage  at  the  load  is  13,200  volts? 
To  what  value  will  the  line  voltage  rise  if  this  load  is  removed  and 
the  field  excitation  of  the  generator  is  unchanged? 

63.  A   1000-kv.-a.,    11,000-volt,    3-phase    generator    delivers 
power  directly  to  a  transmission  line  which  has  a  resistance  of 
8.42  ohms  and  a  reactance  of  6.8  ohms  per  conductor. 


Field  current 

Open-circuit  terminal  voltage 

Rotational  losses 

20 

30 
40 
50 

8,400 
11,000 
12,700 
13,800 

11,600 
19,400 
25,600 

60 

14700 

70 

15.500 

The  armature  has  an  effective  resistance  of  0.94  ohm  per  phase, 
and  the  windings  are  connected  in  Y.  The  field  current  is  supplied 
at  260  volts.  With  the  far  end  of  the  line  short-circuited 
and  with  a  field  current  of  40  amperes  the  line  current  is  115 
amperes . 

If  the  load  at  the  end  of  the  line  requires  940  kw.  at  0.92  power 
factor  what  must  be  the  excitation  of  the  generator  in  order  that 
the  line  voltage  at  the  load  shall  be  11,000  volts? 

64.  A  1000-kw.,  13,800-volt,  3-phase  alternating-current 
generator^ delivers  power  over  a  transmission  line  to  a  synchronous 
motor  load.  The  resistance  and  reactance  of  the  line  are  16.5 
and  17.1  ohms  per  conductor  respectively.  The  armature  of 
the  generator  has  an  effective  resistance  of  2.18  ohms  per  phase. 
The  field  current  is  supplied  with  120  volts. 


POLYPHASE  CIRCUITS 


119 


Field  current 

Open-circuit 
terminal  voltage1 

Core  loss  at 
open  circuit 

50 

80 
110 
140 
180 

8,800 
13,000 
15,600 
17,250 
18,900 

7.5 

16.6 
25.4 
33.5 

With  the  synchronous  motor  running  at  no  load  and  with  under 
excitation  the  line  current  has  its  full-load  value  of  42  amperes, 
the  line  voltage  at  the  motor  is  13,200  volts  and  the  power 
supplied  to  the  motor  is  28.6  kw.  At  this  time  the  field  excita- 
tion of  the  generator  is  161  amperes. 

When  the  motor  load  requires  1000  kw.  what  must  be  the 
excitation  of  the  generator  in  order  that  it  shall  operate  at  unit 
power  factor2  and  the  line  voltage  at  the  load  shall  be  13,200 
volts?  What  is  the  power  factor  of  the  load? 

65.  A  3500-kw.,  10,000-volt,  2-phase  alternating-current 
generator  delivers  power  through  Scott  transformers  to  a  three- 
phase  transmission  line.  With  the  far  end  of  this  line  short- 
circuited  the  output  of  the  generator  is  490  kw.  and  the  armature 
current  and  terminal  voltage  are  220  amperes  and  1630  volts. 
On  open  circuit  the  high-tension  line  voltage  is  33,000  volts  when 
the  terminal  voltage  of  the  generator  is  10,000  volts.  The  arma- 
ture has  an  effective  resistance  of  0.64  ohm  per  phase  The  open- 
and  short-circuit  characteristics  are: 


Field  current 

Open-circuit 
terminal  voltage 

Short-circuit 
armature  current 

100 
200 
250 

7,700 
10,200 
10900 

220 
430 

300 

11,500 

375 

12.200 

If  there  is  a  load  requiring  3200  kw.  at  0.91  power  factor  at  the 
end  of  the  line,  what  must  be  the  excitation  of  generator  in  order 
that  the  line  voltage  at  the  load  shall  be  30,000  volts?  (Use  the 
magnetomotive-force  method  when  dealing  with  the  generator.) 

66.  A  1000-kw.,  13,800-volt,  3-phase  alternating-current 
generator  delivers  power  over  a  transmission  line  which  has  a 

1  The  armature  windings  are  connected  in  Y. 

2The  excitation  of  the  motor  must  also  be  properly  adjusted  for  this  to 
occur. 


120  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 


resistance  of  16.4  ohms  and  a  reactance  of  17.2  ohms  per  con- 
ductor to  a  1340-h.p.  synchronous  motor.  Both  generator  and 
motor  have  their  armatures  connected  in  Y.  This  motor  is 
rated  for  11,000  volts  and  has  an  effective  armature  resistance  of 
0.94  ohms  and  a  synchronous  reactance  of  46  ohms  per  phase. 
The  effective  resistance  and  synchronous  reactance  of  the 
generator  are  respectively  2.18  ohms,  and  82  ohms  per  phase. 
The  rotation  losses  of  the  motor  are  22  kw.  and  may  be  assumed 
constant.  For  certain  excitations  of  generator  and  motor  the 
line  current  and  power  factor  at  the  motor  are  respectively  48 
amperes  and  0.92  (lagging)  when  the  motor  delivers  1140  h.p. 

If  the  excitations  are  unchanged  what  will  be  the  line  current 
and  the  terminal  voltages  of  the  generator  and  motor  if  the  load 
on  the  motor  is  thrown  off? 

67.  A  1500-kv.-a.,  5500-volt  alternator  delivers  energy  to 
a  high-tension  transmission  line  through  step-up  transformers, 
the  low-tension  windings  of  which  are  connected  in  delta  and  the 
high-tension  windings  in  Y.  The  neutrals  of  the  generator  and  of 
the  high-tension  transformer  windings  are  grounded.  The 
transformers  have  a  ratio  of  transformation  of  6.3  to  1.0. 

GENERATOR  CHARACTERISTICS 


Field  current 

Terminal  voltage 
on  open  circuit 

Armature  current 
on  short  circuit 

100 
150 
200 
250 

3500 
5100 
5900 
6500 

200 
300 
400 

300 

6800 

350 

7100 

The  effective  resistance  of  the  armature  is  0.36  ohm  per  phase. 
The  resistance  of  the  field  circuit  is  0.386  ohm.  The  power 
required  to  drive  the  generator  on  open  circuit  with  a  terminal 
voltage  of  5500  volts  is  28.6  kw. 

TRANSFORMER  CHARACTERISTICS 

With  the  high-tension  winding  short-circuited  and  170  volts 
impressed  on  the  low-tension  winding  the  current  supplied  to  a 
transformer  is  91  amperes  and  the  power  is  4.26  kw.  At  no  load 
and  with  5500  volts  impressed  on  the  low-tension  winding  the 
power  is  3.12  kw. 

With  a  balanced  load  of  1360  kw.  at  0.83  power  factor  delivered 


POLYPHASE  CIRCUITS 


121 


to  the  high-tension  line  at  60,000  volts  what  must  be  the  gen- 
erator's excitation?  What  is  the  combined  efficiency  of  the  gen- 
erator and  the  transformers  at  this  load?  If  this  load  were 
removed  what  would  be  the  high-tension  line  voltage  if  the 
excitation  of  the  generator  were  unchanged?  Use  (a)  the  syn- 
chronous impedance  method  and  (b)  the  magnetomotive-force 
method  when  calculating  the  field  current  of  the  generator. 

68.  A  1000-kv.-a.,  2400-volt  alternator  delivers  energy  to  a 
high-tension  transmission  line  through  step-up  transformers  both 
the  low-  and  high-tension  windings  of  which  are  connected  in  Y. 
The  armature  windings  of  the  generator  which  are  connected  in 
delta  have  an  effective  resistance  of  0.20  ohm  per  phase.  The 
resistance  of  the  field  circuit  is  0.427  ohm. 

GENERATOR  CHARACTERISTICS 


Field 
current 

Open-circuit 
terminal  voltage 

Terminal  voltage 
la  =  139.     P.F.=Q 

Core  loss  on 
open  circuit 

60 

1470 

7  6 

100 

2220 

17  9 

140 
180 
220 

2700 
2980 
3180 

1920 
2370 
2630 

28.5 
37.0 

The  friction  and  windage  loss  is  8.2  kw. 

TRANSFORMER  CHARACTERISTICS 

The  transformers  have  a  ratio  of  transformation  of  1390:12,700 
volts.  With  the  low-tension  winding  short-circuited  and  with 
332  volts  impressed  on  the  high-tension  winding  the  current  is 
26.2  amperes  and  the  power  is  3120  watts.  The  core  loss  at  the 
rated  voltage  is  2220  watts. 

The  heaviest  load  that  the  high-tension  line  requires  is  1000  kw. 
at  0.88  power  factor  and  the  necessary  line  voltage  is  24,500  volts. 
What  must  be  the  terminal  voltage  of  the  exciter  for  this  load? 
What  is  the  combined  efficiency  of  the  generator  and  the  trans- 
formers at  this  load? 

69.  A  1500-kv.-a.,  5500-volt  alternator  delivers  energy  to  a 
high-tension  transmission  line  through  three  step-up  transformers, 
the  low-tension  windings  of  which  are  connected  in  delta  and  the 
high-tension  windings  in  Y.  The  neutrals  of  the  generator  and 
of  the  high-tension  transformer  windings  are  grounded.  The 
transformers  have  a  ratio  of  transformation  of  3.2  to  1.0.  The 
calculated  resistance  and  reactance  of  the  high-tension  line  are 


122  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 


48.6  ohms  and  59.4  ohms  per  conductor  respectively.  The  gen- 
erator has  an  effective  armature  resistance  of  0.36  ohm  per  phase. 
The  open-  and  short-circuit  characteristic  data  for  the  generator 


are: 


Field 
current 

Terminal  voltage 
on  open  circuit 

Armature  current 
on  short  circuit 

100 
150 
200 

3500 
5100 
5900 

200 
300 

250 
300 
350 

6500 
6800 
7100 

With  the  far  end  of  the  transmission  line  short-circuited  the 
generator  supplies  a  current  of  165  amperes  at  a  terminal  potential 
of  860  volts,  and  the  power  delivered  to  the  transformers  and  line 
is  146  kw. 

If,  when  the  far  end  of  this  transmission  line  delivers  a  balanced 
load  of  1450  kw.  at  a  power  factor  of  0.93,  the  line  potential  differ- 
ence at  the  load  is  30,000  volts,  to  what  value  would  this  voltage 
rise  if  the  load  were  removed?  Use  what  you  consider  the  most 
exact  method  of  calculation. 

70.  A  1000-kv.-a.,  13,800-volt  generator  delivers  power  over 
a  transmission  line  and  through  step-down  transformers  which  are 
connected  in  delta  on  both  the  high-  and  low-tension  sides.  The 
armature  windings  of  the  generator,  which  are  connected  in  Y, 
have  an  effective  resistance  of  2.18  ohms  per  phase.  The  resist- 
ance of  the  field  circuit  is  0.541  ohm. 

GENERATOR  CHARACTERISTICS 


Field 
current 

Terminal  voltage 
on  open  circuit 

Terminal  voltage 
la  =  42    P.F.=Q 

Core  loss  on 
open  circuit 

50 

8,800 

7  5 

80 

13,000 

16  6 

110 
140 
180 

15,600 
17,250 
18,900 

10,750 
13,250 
15,600 

25.4 
33.5 

The  friction  and  windage  loss  is  7.9  kw. 

TRANSFORMER  CHARACTERISTICS 

Each  transformer  has  a  ratio  of  transformation  of  425:13,200 
volts.  With  the  low-tension  winding  short-circuited  and  with 
1100  volts  impressed  on  the  high-tension  winding  the  current 


POLYPHASE  CIRCUITS 


123 


is  25.2  amperes  and  the  power  is  3070  watts.  The  core  loss 
at  the  rated  voltage  is  2130  watts. 

The  calculated  resistance  and  reactance  of  the  transmission 
line  are  15.7  ohms  and  16.2  ohms  per  conductor. 

With  the  low-tension  windings  of  the  transformers  short-cir- 
cuited what  must  be  the  terminal  voltage  of  the  generator  in 
order  that  it  will  deliver  its  full-load  current?  When  the  trans- 
formers deliver  a  load  of  1000  kv.-a.  at  0.90  power  factor  and  a 
line  voltage  of  420  volts  what  is  the  necessary  excitation  of  the 
generator,  and  what  is  the  combined  efficiency  of  the  generator, 
line  and  transformers? 

71.  An  850-kv.-a.,  11,000-volt,  Y-connected  generator  delivers 
energy  over  a  transmission  line  and  through  transformers,  that 
are  connected  in  Y  on  the  high-tension  and  in  delta  on  the  low- 
tension  sides,  to  a  load  of  induction  motors  and  a  synchronous 
motor.  The  resistance  and  reactance  of  the  line  are  11.4  ohms 
and  10.6  ohms  per  conductor  respectively.  The  transformers  are 
each  rated  to  deliver  300  kw.  with  a  ratio  of  transformation  of 
6600:660  volts. 

GENERATOR  CHARACTERISTICS 


Field 
current 

Open-circuit 
phase  voltage 

Short-circuit 
armature  current 

25 
50 

2200 
4070 

72  . 

75 
100 

5500 
6270 

125 
150 
175 

6930 
7370 
7700 

The  ohmic  resistance  of  the  armature  between  terminals  is 
3.32  ohms  at  the  running  temperature.  The  effective  resis- 
tance is  1.45  times  the  ohmic  resistance.  The  field  resistance  is 
0.723  ohm.  The  core  and  friction  losses  are  26  kw.  at  the  rated 
voltage. 

TRANSFORMER  CHARACTERISTICS 

With  the  low-tension  winding  short-circuited  and  with  240 
volts  impressed  on  the  high-tension  winding  the  current  is  47 
amperes  and  the  power,  2.8  kw.  The  core  loss  at  the  rated  vol- 
tage is  2.6  kw. 

The  induction  motors  take  a  constant  load  of  220  kw.  at  0.83 


124  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

power  factor.  The  synchronous  motor  takes  a  constant  Joad  of 
460  kw.,  and,  with  the  greatest  allowable  field  excitation,  its  line 
current  is  610  amperes  and  the  terminal  voltage  is  boosted  to 
650  volts. 

Calculate  the  terminal  voltage  of  the  generator's  exciter  and 
the  combined  efficiency  of  the  generator,  line  and  transformers. 

Use  what  you  consider  the  most  exact  method. 

72.  A   constant  induction  motor   load  taking   2500   kw.   at 
0.91  power  factor  and  with  a  line  voltage  of  13,200  volts  is  at 
the  end  of  a  short  transmission  line.     For  this  load  the  efficiency 
of  transmission  is  91.3  per  cent,  and  the  voltage  regulation  of  the 
line  is  11.2  per  cent.     A  synchronous  motor  of  suitable  capacity 
is  added  at  the  load  so  that  when  running  light  with  full-load 
current  the  resultant  power  factor  at  the  load  is  increased  to 
unity.     Assume  that  the  efficiency  of  this  synchronous  motor 
at  full  load  and  unit  power  factor  is  0.93.     The  voltage  at  the 
load  is  maintained  constant. 

What  is  the  necessary  capacity  of  the  synchronous  motor? 
What  are  the  efficiency  of  transmission  and  the  voltage  regula- 
tion of  the  line  after  the  synchronous  motor  is  added? 

73.  An  induction  motor  load  at  the  end  of  a  three-phase  25-cycle 
transmission  line   takes   7600  kw.  at   0.912   power  factor  and 
with  a  line  voltage  of  11,000  volts.     The  resistance  and  induc- 
tance of  the  line  are  respectively  1.4  ohms  and  6.5  millihenrys  per 
conductor.     A   synchronous   motor,    running  light   and   taking 
full-load  current,  is  added  at  the  generating  station  to  improve 
the  power  factor  and  thus  increase  the  capacity  of  the  line  at  the 
load.     Assume  that  the  full-load  efficiency  of  this  motor  is  94 
per  cent,  when  operating  at  unit  power  factor.     The  voltage 
at  the  load  is  maintained  constant.     Induction  motors,  operating 
at  the  same  power  factor  as  do  the  others,  are  added  at  the  load 
and  the  synchronous  motor  is  adjusted  so  that  the  generating 
station  operates  at  unit  power  factor.     For  the  same  line  cur- 
rent in  the  station  as  was  required  before  the  addition  of  the 
synchronous   motor   calculate  the  permissible  increase  in  the 
induction  motor- load.     What  is  the  kilovolt-ampere  capacity  of 
the  necessary  synchronous  motor?     What  is  the  line  voltage  at 
the  generating  station  before  and  after  the  synchronous  motor 
is  added? 

74.  An  induction  motor  load  taking  5400  kw.  at  0.914  power 
factor  and  with  a  line  voltage  of  13,200  volts  is  at  the  end  of  a 


POLYPHASE  CIRCUITS  125 

three-phase  transmission  line,  which  has  a  resistance  of  2.38  ohms 
and  an  inductance  of  7.6  millihenrys  per  conductor.  The  full- 
load  capacity  of  the  generating  station  is  320  amperes  per  line. 
A  synchronous  motor  is  to  be  added  at  the  load  both  to  compensate 
for  power  factor  and  to  supply  additional  power.  Assume  that 
the  efficiency  of  this  synchronous  motor  and  its  exciter  is  92  per 
cent,  at  full  load  and  with  unit  power  factor.  The  voltage  at 
the  load  is  maintained  constant.  The  frequency  is  60  cycles. 

Calculate  the  kilovolt-ampere  capacity  of  the  synchronous  motor 
so  that  the  generating  station  can  deliver  its  full-load  current  at 
unit  power  factor.  What  additional  power  can  the  synchronous 
motor  supply?  At  what  power  factor  does  the  synchronous 
motor  operate?  What  is  the  necessary  line  voltage  at  the 
generating  station  before  and  after  the  synchronous  motor 
is  added? 

75.  An  induction  motor  load  taking  26,000  kw.  at  0.906 
power  factor  and  with  a  line  voltage  of  6600  volts  is  operated  at 
the  end  of  a  high-tension  transmission  line.  At  the  ends  of  the  line 
there  are  step-up  and  step-down  transformers,  which  have  the 
same  ratio  of  transformation.  The  total  equivalent  resistance  and 
reactance  of  the  line  and  transformers  referred  to  the  low-tension 
sides  are  0.431  ohm  and  0.986  ohm  at  25  cycles  respectively.  Syn- 
chronous motors,  running  light,  but  taking  their  full-load  current, 
are  added  at  the  load  to  improve  the  power  factor  and  thus 
increase  the  capacity  of  the  generating  station.  Assume  that  the 
full-load  efficiency  of  the  synchronous  motors  and  their  exciters 
is  0.92  at  unit  power  factor.  The  voltage  at  the  load  is  main- 
tained constant.  Induction  motors  operating  at  the  same  power 
factor  as  do  the  others  are  added  at  the  load  and  the  synchronous 
motors  are  adjusted  so  that  the  resultant  power  factor  of  the  load 
is  increased  to  unity. 

For  the  same  line  current  as  required  before  the  addition  of 
the  synchronous  motors  calculate  the  permissible  increase  in  the 
induction  motor  load.  What  is  the  necessary  kilovolt-ampere 
capacity  of  the  synchronous  motors?  At  what  power  factor  was 
the  generating  station  operating  before  and  after  the  synchronous 
motors  were  added?  What  was  the  line  voltage  at  the  generating 
station  before  and  after  the  synchronous  motors  were  added? 


CHAPTER  VII 
NON-SINUSOIDAL  WAVES 

1.  The  equations  for  the  open-circuit   phase   voltages   of   a 
three-phase,  Y-connected;  alternating-current  generator  are  • 

61  =  180  sin  0^+60  sin  3  cut 

2n  2r 

e2=180  sin  (cut  —  ^~)+60  sin 


-x 
e3=180  sin  (art—  «-)  60  sin 

What  is  the  equation  of  the  line  voltage,  612?  What  would  a 
voltmeter  indicate  when  connected  across  one  phase?  When 
connected  across  the  line  terminals? 

2.  The  equations  for  the  open-circuit  phase  voltages  of  a  three- 
phase,  Y-connected,  alternating-current  generator  are  : 

ei  =  5300  sin  ^+1200  sin  3  cut 

2x 
62  =  5300  sin  (orf—  -)  +  1200  sin 


63  =  5300  sin  (orf-y)+1200  sin  3(oj<-y). 

What  is  the  equation  for  the  line  voltage  6i2?  What  would  a 
voltmeter  indicate  when  connected  across  one  phase?  When 
connected  across  the  line  terminals? 

3.  The  equation  for  the  voltage  between  line  and  neutral  of 
a  four-phase  generator  is: 

61  =  1600  sin  6^+500  sin  3  wt. 

What  is  the  equation  for  the  voltage  between  adjacent  line 
terminals?  What  would  a  voltmeter  indicate  when  connected 
between  line  and  neutral?  Between  adjacent  lines? 

4.  The  equation  for  the  voltage  between  line  and  neutral  of  a 
four-phase  generator  is: 

61  =  1600  sin  itrt+400  sin  5  wt. 

What  is  the  equation  for  the  voltage  between  adjacent  line  ter- 

126 


NON-SINUSOIDAL  WAVES  127 

minals?     What  would   a  voltmeter   indicate   when   connected 
between  line  and  neutral?     Between  adjacent  lines? 

5.  The  phase  voltage  of  a  three-phase,  Y-connected,  alternat- 
ing-current generator  is  3750  volts.     This  consists  of  a  funda- 
mental and  a  third  harmonic  which  is  30  per  cent,  of  the  funda- 
mental.    What  is  the  line  voltage? 

6.  The  phase  voltage  of  a  three-phase,  Y-connected,  alternat- 
ing-current generator  is  2900  volts.     This  consists  of  a  funda- 
mental and  a  fifth  harmonic  which  is  25  per  cent,  of  the  funda- 
mental.    What  is  the  line  voltage? 

7.  The  phase  voltage  of  a  3-phase,  Y-connected,  alternating- 
current  generator  is  3980  volts,  while  the  line  voltage  is  6600 
volts  at  the  same  time.     If  it  is  assumed  that  the  phase  voltage  con- 
tains no  harmonic  higher  than  the  seventh,  what  is  the  magnitude  of 
the  third  harmonic  in  the  phase  voltage? 

8.  The  phase  voltage  of  a  three-phase,  Y-connected,  alternat- 
ing-current generator  is  3010  volts  while  the  line  voltage  (at  the 
same  time)  is  5000  volts.     What  is  the  greatest  value  that  a  third 
harmonic  component  of  the  phase  voltage  can  have? 

9.  The  phase  voltage  of  a  three-phase,  Y-connected,  alter- 
nating-current generator  is  138  volts  while  at  the  same  time  the 
line  voltage  is  230  volts.     If  the  phases  of  this  generator  were  con- 
nected in  delta,  what  would  be  the  unbalanced  voltage  tending  to 
set  up  a  circulating  current  in  the  windings? 

10.  Two  transformers  are  arranged  after  the  Scott  method 
of  connection  to  transform  power  from  two-phase  to  three-phase. 
Their  ratios  of  transformation  are  1:10  and  1:8.66. 

If  there  are  impressed  on  the  two-phase  side  electromotive 
forces  whose  equations  are 

6i  =  1550  sin  a)t+ 500  sin  3  cot, 

and      62  =  1550  sin  (cot  —  ^+500  sin  3(cut  —  ^), 

what  are  the  equations  for  the  line  voltages  on  the  three-phase 
side? 

11.  In  problem  10  if  there  are  impressed  on  the  two-phase  side 
electromotive  forces  whose  equations  are, 

61  =  1550  sin  &>£+300  sin  5  cot, 
and      e2  =  1550  sin  (cut  —  ^+300  sin  5(tot  —  ^) 


128  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

what  are  the  equations  for  the  line  voltages  on  the  three-phase 
side? 

12.  In  problem  10  there  are  impressed  on  the  two-phase  side 
equal  electromotive  forces  which  contain  a  third  and  a  fifth  har- 
monic.    The  effective  value   of  these  electromotive  forces    is 
1100  volts  and  the  harmonics  are  respectively  0.3  and  0.2  of  the 
fundamental.     What  are  the  effective  values  of  the  three-phase 
line  voltages?     What  per  cents,  of  the  fundamental  components 
are  each  of  the  harmonics  in  these  line  voltages? 

13.  The  line  voltage  and  the  voltage  to  neutral  of  a  balanced 
three-phase  circuit  are  respectively  230  volts  and  139  volts.     The 
voltage  to  neutral  contains  a  fundamental  and  a  third  harmonic 
only.     Three  equal  resistance  units  of  10  ohms  each  are  connected 
in  Y  across  the  lines  of  this  circuit. 

(a)  What  is  the  line  current?     What  are  the  line  and  neutral 
currents  when  the  neutral  point  of  the  resistance  units  is  connected 
to  the  neutral  of  the  circuit. 

(b)  If  the  power  is  measured  by  the  two-wattmeter  method 
what  would  be  the  indicated  power  before  and  after  the  connection 
to  the  neutral  is  made?     If  a  wattmeter  were  connected  in  the 
circuit  with  its  current  coil  in  the  neutral  and  its  potential  coil 
between  the  neutral  and  one  of  the  lines,  what  would  this  watt- 
meter indicate  in  the  second  case? 

14.  The  voltage  to  neutral  of  a  balanced  three-phase  circuit 
is  134  volts,  and  it  contains  a  fundamental  and  a  fifth  harmonic 
which  is  0.2  of  the  fundamental.     Three  equal  resistance  units  of 
10  ohms  each  are  connected  in  Y  across  the  lines  of  this  circuit. 
What  is  the  line  current?     If  the  neutral  point  of  the  resistance 
units  and  the  neutral  of  the  circuit  are  connected  what  is  the  cur- 
rent in  the  neutral  conductor?     What  is  the  total  power  supplied 
in  each  case? 

15.  The  line  voltage  and  the  voltage  to  neutral  of  a  balanced 
three-phase,  60-cycle  circuit  are  respectively  230  volts  and  139 
volts.     The  voltage  to  neutral  contains  a  fundamental   and  a 
third  harmonic  only.     Three  equal  air-core  reactors,  each  having  a 
resistance  of  5  ohms  and  an  inductance  of  0.015  henry,  are  con- 
nected in  Y  across  the  lines  of  this  circuit. 

(a)  What  is  the  line  current?     What  are  the  line  and  neutral 
currents  when  the  neutral  point  of  the  reactors  is  connected  to 
the  neutral  of  the  circuit? 

(b)  If  the  power  is  measured  by  the  two-wattmeter  method 


NON-SINUSOIDAL  WAVES  129 

what  is  the  indicated  power  before  and  after  the  connection  to  the 
neutral  is  made?  If  a  wattmeter  is  connected  in  the  circuit  with 
its  current  coil  in  the  neutral  and  its  potential  coil  between  the 
neutral  and  one  of  the  lines,  what  would  this  wattmeter  indicate 
in  the  second  case? 

16.  The  voltage   to   neutral   of  a  balanced,  three-phase,  60- 
cycle  circuit  is  134  volts,  and  it  contains  a  fundamental  and  a  fifth 
harmonic  which  is  0.2  of  the  fundamental  component.     Three 
equal  air-core  reactors,  each  having  a  resistance  of  5.0  ohms  and 
an  inductance  of  0.015  henry,  are  connected  in  Y  across  the  lines 
of  this  circuit.     What  is  the  line  current?     If  the  neutral  point  of 
these  reactors  and  the  neutral  of  the  circuit  are  connected  what  is 
the  current  in  the  neutral  conductor?     What  is  the  total  power 
supplied  in  each  case? 

17.  The  line  voltage  and  the  voltage  to  neutral  of  a  balanced 
three-phase,  60-cycle  circuit  are  230  volts  and  139  volts  respect- 
ively.    The  voltage  to  neutral  contains  a  fundamental  and   a 
third  harmonic  only.     Three  equal  impedance  units,  each  con- 
sisting of  a  resistance  of  5   ohms  in  series  with  a  capacity  of  25 
microfarads  are  connected  in  Y  across  the  lines  of  this  circuit. 

(a)  What  is  the  line  current?     What  are  the  line  and  neutral 
currents  when  the  neutral  point  of  this  load  is  connected  to  the 
neutral  point  of  the  circuit? 

(b)  If  the  power  is  measured  by  the  two-wattmeter  method 
what  is  the  indicated  power  before  and  after  the  connection  to  the 
neutral  is  made?     If  a  wattmeter  is  connected  in  the  circuit  with 
its  current  coil  in  the  neutral  and  its  potential  coil  between  the 
neutral  and  one  of  the  lines,  what  will  it  indicate  in  the  second 
case? 

18.  The  voltage  to  neutral  of  a  balanced  three-phase,  60-cycle 
circuit  is  134  volts,  and  it  contains  a  fundamental  and  a  fifth 
harmonic  which  is  0.2  of  the  fundamental  component.     Three 
equal   impedance   units,    each   consisting   of  a  resistance   of  5 
ohms  in  series  with  a  capacity  of  25  microfarads,  are  connected  in 
Y  across  the  lines  by  this  circuit.     What  is  the  line  current?     If 
the  neutral  point  of  this  load  is  connected  to  the  neutral  point  of 
the  circuit  what  is  the  current  in  the  neutral  conductor?     What 
is  the  total  power  supplied  in  each  case? 

19.  Three  unequal  lamp    loads   are    connected   between   the 
lines  and  neutral  of  a  balanced  three-phase  circuit.     The  line 
voltage  is  230  volts  and  the  voltage  to  neutral  is  known  to  contain 


130  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

a  third  harmonic  which  is  0.3  of  the  fundamental  component.  The 
resistances  of  the  lamp  loads  are  5,  8,  and  10  ohms.  What  are 
the  line  currents,  the  neutral  current,  and  the  total  power 
supplied? 

20.  Three  equal  lamp  loads  are  connected  in  Y  across  the  lines 
of  a  balanced  three-phase  circuit.     The  line  voltages  are  230 
volts  and  the  voltage  from  line  to  neutral  is  139  volts.     The  latter 
voltage  is  known  to  contain  harmonics.     The  resistance  of  each  of 
the  lamp  loads  is  5.0  ohms.     What  will  be  the  effect  on  the  line 
current  if  the  neutral  point  of  the  load  is  connected  through 
a  resistance  of  2  ohms  to  the  neutral  of  the  circuit?     By  what 
amount  is  the  power  supplied  to  the  lamps  increased?     What  is 
the  loss  in  the  2-ohm  resistance? 

21.  The  voltages  from  the  lines  to  the  neutral  conductor  of  a 
balanced  three-phase  circuit  are  each  140  volts.     Three  equal 
resistance  units  of  10  ohms  each  are  connected  in  Y  across  this 
circuit.     A  voltmeter  connected  between  the  neutral  point  of 
this  load  and  the  neutral  conductor  indicates  40  volts.     What 
power  will  these  resistance  units  take  if  the  neutral  point  of  the 
load  is  directly  connected  to  the  neutral  conductor?     What  will 
be  the  current  in  the  neutral  conductor? 

22.  The  line  voltages  of  a  three-phase  circuit  are  equal  and 
maintained  constant.     The  voltages  from  the  lines  to  the  neutral 
conductor,  which  are  known  to  contain  third  harmonics,  are  also 
maintained  constant  and  are  each  equal  to  140  volts.     Three 
equal  resistance  units  of  10  ohms  each  are  connected  in  Y  across 
this  circuit.     The  measured  voltage  from  the  neutral  of  this 
load  to  the  neutral  conductor  is  40  volts.     What  current  will 
exist  in  the  neutral  conductor  if  a  resistance  unit  of  5  ohms  is 
connected  between  these  two  points?     What  is  the  effective 
voltage  across  each  of  the  equal  resistance  units  before  and  after 
this  additional  resistance  is  inserted  in  the  circuit? 

23.  Three  hypothetical  impedance  units,  a  resistance,  a  re- 
actor, and  a  condenser  are  so  constructed  that  at  60  cycles  the 
values  of  their  impedances  may  be  represented  in  the  complex 
notation  by:     2i  =  10+jQ,    z2  =  0+jlO   and   z2  =  0-j!0.     These 
impedance  units  are  connected  between  the  mains  and  neutral 
of  a  balanced  three  phase,  60-cycle  circuit      The  line  voltages  are 
each  230  volts,  and  the  voltages  between  the  lines  and  the  neutral 
are  equal  and  consist  of  a  fundamental  and  a  third  harmonic 
which  is  0.3  of  the  fundamental  component.     The  first  impedance 


NON-SINUSOIDAL  WAVES  131 

unit  is  connected  from  line  (1),  the  second,  from  line  (2),  and  the 
third,  from  line  (3)  to  the  neutral. 

If  the  cyclic  order  of  the  line  voltages  is  such  that  the  funda- 
mental component  of  Viz  leads  Vzs  by  120  degrees  what  are  the 
line  currents?  What  is  the  neutral  current?  What  is  the  total 
power  supplied? 

24.  In  problem  23  if  the  cyclic  order  of  the  line  voltages  is  such 
that  the  fundamental  component  of  Viz  lags  Vzs  by  120  degrees 
what  are  the  line  currents?     What  is  the  neutral  current?     What 
is  the  total  power  supplied? 

25.  Three  hypothetical  impedance  units,  a  resistance,  a  react- 
ance, and  a  condenser  are  so  constructed  that  at  60  cycles  their 
values  may  be  presented  in  the  complex  notation  by:  z\  — 10+ JO, 
22  =  0+jlO,  23  =  0— jlO.     These  impedance  units  are  connected 
between  the  mains  and  neutral  of  a  balanced  three-phase,  60- 
cycle  circuit.     The  line  voltages  are  each  230  volts,  and  the  vol- 
tages between  the  lines  and  the  neutral  are  equal  and  consist  of  a 
fundamental    and    a    fifth    harmonic    which    is    0.25    of    the 
fundamental    component.     The   first   impedance   is   connected 
from  line  (1),  the  second,  from  line  (2),  and  the  third,  from  line 
(3)  to  the  neutral.     If  the  cyclic  order  of  the  line  voltages  is 
such  that  the  fundamental  component  of  Viz  leads  Vzs  by  120 
degrees  what  are  the  line  currents?     What  is  the  neutral  current? 
What  is  the  total  power  supplied? 

26.  In  problem  25  if  the  cyclic  order  of  the  line  voltages  is 
such  that  the  fundamental  component  of  Viz  lags  Vzs  by  120 
degrees  what  are  the  line  currents?    What  is  the  neutral  current? 
What  is  the  total  power  supplied? 

27.  Three   hypothetical   impedance   units,    a   resistance,    a 
reactor,  and  a  condensor  are  so  constructed  that  at  60  cycles 
their  values  may  be  represented  in  the  complex  notation  by: 

zi  =  10-f-jO,  22  =  0+jlO,  z3  =  0— jlO.  These  impedance  units  are 
connected  in  delta  across  the  lines  of  a  balanced  three-phase, 
60-cycle  circuit.  The  line  voltages  are  each  230  volts  and  con- 
sist of  a  fundamental  and  a  fifth  harmonic  which  is  0.25  of  the 
fundamental  component.  The  first  impedance  is  connected 
between  lines  1-2,  the  second,  between  lines  2-3,  and  the  third, 
between  lines  3-1. 

If  the  cyclic  order  of  the  line  voltages  is  such  that  the  funda- 
mental component  of  Viz  leads  Vzs  by  120  degrees  what  are  the 
line  currents? 


132  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

28.  In  problem  27  if  the  cyclic  order  of  the  line  voltages  is  such 
that  the  fundamental  component  of  Viz  lags  V%$  by  120  degrees 
what  are  the  line  currents? 

29.  Three  equal  resistance  units  of  50  ohms  each  are  connected  in 
Y  across  the  mains  of  a  balanced  three-phase,  230-volt  circuit.    An 
air-core  reactor  which  has  a  resistance  of  0.5  ohm  and  an  induc- 
tance of  5  millihenrys  is  connected  from  neutral  point  of  this 
load  through  a  switch  to  the  neutral  conductor  of  the  circuit.    The 
voltage  across  the  open  switch  is  40  volts.     There  is  no  higher 
harmonic  than  a  third  present. 

(a)  With  the  switch  open  what  is  the  line  current?     What  is 
the  total  power  supplied? 

(b)  With  the  switch  closed  what  is  the  line  current?     What  is 
the  neutral  current?     What  is  the  total  power  supplied?     What 
are  the  voltages  across  the  resistance  units  before  and  after  the 
switch  is  closed?     What  is  the  voltage  across  the  reactor  after 
the  switch  is  closed? 

30.  In  problem  29  if  the  air-core  reactor  is  replaced  by  a  con- 
denser which  has  a  capacity  of  50  microfarads  what  are  the  line 
currents  before  and  after  the  switch  is  closed?     What  is  the 
neutral  current  after  the  switch  is  closed?     What  is  the  total 
power  supplied  with  the  switch  open?     With  the  switch  closed? 
What  are  the  voltages  across  the  resistance  units  before  and  after 
the  switch  is  closed?     What  is  the  voltage  across  the  condenser 
after  the  switch  is  closed? 

31.  The  voltages  between  the  lines  and  neutral  conductor  of  a 
balanced  three-phase,  60-cycle  circuit  are  each  140  volts.     Three 
equal  condensers  each  of  5  microfarads'  capacity  are  connected 
in  Y  across  the  lines  of  this  circuit,  and  from  their  common  junc- 
tion a  non-inductive  resistance  unit  of  10  ohms'  resistance  is  con- 
nected through  a  switch  to  the  neutral  conductor.     The  voltage 
across  the  open  switch  is  45  volts.     There  is  no  higher  harmonic 
than  a  third  present,     (a)  With  the  switch  open  what  is  the  line 
current?     What  is  the  total  power  supplied?     (b)  With  the  switch 
closed  what  is  the  line  current?     What  is  the  neutral  current? 
What  is  the  total  power  supplied? 

32.  In  problem    31  if    the    non-inductive    resistance  unit  is 
replaced  by  an  air-core  reactor  which  has   a   resistance  of  0.5 
ohm  and  an  inductance  of  5  millihenrys  what  are  the  line  cur- 
rents before  and  after  the  switch  is  closed?     What  is  the  neutral 
current  after  the  switch  is  closed?     What  is  the  total  power  sup- 


NON-SINUSOIDAL  WAVES  133 

plied  with  the  switch  closed?  What  are  the  voltages  across  the 
condensers  before  and  after  the  switch  is  closed?  What  is  the 
voltage  across  the  reactor  after  the  switch  is  closed? 

33.  Three  equal  lamp  loads  each  of  which  has  a  resistance  of 
5  ohms  are  connected  in  Y  across  the  lines  of  a  balanced  230-volt, 
3-phase  circuit.     Due  to  harmonics  the  voltages  from  the  lines  to 
the  neutral  conductor  of  the  circuit  are  each  140  volts.     The 
power  supplied  is  measured  by  the  two-wattmeter  method. 

(a)  What  are  the  line  currents  and  the  wattmeter  readings? 

(b)  If  the  neutral  point  of  the  load  is  connected  to  the  neutral 
conductor  what  will  the  wattmeters  read?     What  is  the  power 
supplied?     Compare  the  true  power  factor  of  this  load  with  that 
calculated  from  the  wattmeter,  voltmeter,  and  ammeter  readings. 

34.  Three  unequal  lamp  loads  are  connected  between  the  lines 
and  the  neutral  conductor  of  a  balanced  230-volt,  3-phase  circuit 
Due  to  third  and  fifth  harmonics,  which  are  respectively  0.3  and. 
0.25  of  the  fundamental,  the  voltages  from  the  lines  to  the  neutral 
conductor  are  each  140  volts.     The  effective  line  currents  are 
20  amperes,  30  amperes  and  40  amperes  respectively.     What  is 
the  current  in  the  neutral  conductor? 

35.  In  problem  34  if  fuses  in  the  third  line  and  in  the  neutral 
conductor  blow,  what  current  will  the  lamp  loads  take,  and  what 
will  be  the  voltage  across  each  of  them?     Assume  that  the  resist- 
ance of  each  lamp  circuit  is  constant. 

36.  By  mistake  three  equal  air-core  impedance  units  are  con- 
nected in  delta  across  two  of  the  mains  and  the  neutral  conductor 
of  a  three-phase,  60-cycle  circuit.     Each  of  these  impedance  units 
has  a  resistance  of  5  ohms  and  reactance  of  2  ohms  at  60  cycles. 
The  voltages  between  the  mains  and  neutral  conductor  are  each 
140  volts,  and  the  voltage  between  any  two  of  the  mains  is  230 
volts.     Assume  that  there  are  no  harmonics  higher  than  the  third 
present.     What  are  the  currents  in  each  of  the  impedance  units? 
What  are  the  currents  in  the  mains  and  in  the  neutral  conductor? 

37.  Two  suitable  transformers  are  arranged  after  the  Scott 
method  of  connection  to  transform  from  2-phase  to  3-phase. 
Each  of  the  two-phase  line  voltages  is  2200  volts  and  consists  of 
a  fundamental  and  a  third  harmonic  which  is  30  per  cent,  of  the 
fundamental.     The  3-phase  line  voltages  each  have  an  effective 
value  of  230  volts.     The  frequency  of  the  fundamental  is  60 
cycles.     Neglect  the  resistance  and  the  leakage  reactance  of  the 
transformers. 


134  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

(a)  If  three  equal  lamp  loads,  each  of  which  has  a  resistance  of 
5  ohms,  are  connected  in  delta  across  the  3-phase  lines  what 
current  does  each  take? 

(b)  If  three  equal  lamp  loads,  each  of  which  has  a  resistance  of 
3  ohms,  are  connected  in  Y  across  the  3-phase  lines  what  current 
does  each  take? 

38.  In  problem  37  three  equal  air-core  reactors,  each  of  which 
has  a  resistance  of  5  ohms  and  an  inductance  of  7.5  mil-henry s, 
are  connected  in  delta  across  the  3-phase  lines.     What  current 
does  each  take?     What  is  the  line  current? 

(b)  If  these  reactors  are  connected  in  Y  what  current  will 
each  take? 

39.  In  problem  37    three   equal  impedance  units,  each  con- 
sisting of  a  resistance  of  20  ohms  in  series  with  a  condenser 
of  50  microfarads'  capacity,  are  connected  in  delta  across  the  3- 
phase  lines.     What  current  does  each  unit  take?     What  is  the 
line  current? 

(b)  If  these  impedance  units  are  connected  in  Y  what  current 
will  each  take? 

40.  In  problem  37  three  hypothetical    impedance   units   are 
connected  in  delta  across  the  3-phase  lines.     The  values  of  these 
units  at  60  cycles  may  be  represented  in  the  complex  notation  by : 
*i  =  10+jO,    2fe  =  0+jlO,   z3  =  0-jlO.     What   current   does   each 
unit  take?     What  are  the  line  currents? 

41.  In  problem  37  the  smaller   transformer   is  tapped   at  a 
point  two-thirds  from  its  line  terminal  to  give  a  neutral  on  the 
3-phase  side.     The  three  hypothetical  impedance  units  described 
in  problem  40  are  connected  in  Y  between  the  lines  and  neutral. 
What  are  the  line  currents?     What  is  the  neutral  current? 

42.  A  3-phase,  60-cycle  alternating-current  generator  supplies 
power  to  a  transmission  line  through  step-up  transformers  whose 
low  tension  windings  are  connected  in  Y  and  whose  high-tension 
windings  are  connected  in  delta.     These  transformers  are  rated  at 
3333  kv.-a. ,  and  have  a  voltage  ratio  of  3,800  to  80,000  volts.     With 
the  high-tension  winding  short-circuited  and  with  162  volts  at  60 
cycles  impressed  on  the  low-tension  winding  full-load  current 
exists  in  each  winding  and  18.85  kw.  is  supplied.     The  neutral 
points  of  both  the  generator  and  the  low-tension  windings  of  the 
transformers  are  connected  to  the  same  ground  bus.     When  the 
transformers  are  delivering  no  load  on  the  high-tension  side  and 
the  excitation  of  the  generator  is  adjusted  so  that  its  terminal 


NON-SINUSOIDAL  WAVES  135 

voltage  is  6600  volts  the  voltage  from  line  to  neutral  is  found  to 
be  3850  volts.  A  third  harmonic  in  the  phase  voltage  is  sus- 
pected, and  the  oscillograph  shows  that  one  with  a  magnitude  of 
14  per  cent,  of  the  fundamental  does  exist. 

What  is  the  copper  loss  in  each  transformer  when  there  is  no 
load  delivered  to  the  high-tension  line.  Compare  this  with  the 
full-load  copper  loss  that  would  be  produced  in  the  transformers 
if  their  neutral  point  was  not  grounded. 

43.  A  1000-kw.,  6-phase,  25-cycle,  600-volt  rotary  converter  is 
supplied  with  power  from  a  3-phase,   13,200-volt  transmission 
line.     The  high-tension  windings  of  the  transformers  are  con- 
nected  in  delta  and  the  low-tension  windings  are  connected  to 
diametrical  points  of  the  armature  of  the  rotary.     With  the  low- 
tension  winding  of  a  transformer  short-circuited  and  with  345 
volts  at  25  cycles  impressed  on  the  high-tension  winding,  full- 
load  current,  or  25.2  amperes  exists  in  the  winding  and  3070 
watts  is  supplied.     When  the  rotary  is  delivering  about  full  load 
and  the  excitation  is  adjusted  for  near  unit  power  factor  receive 
the  transformer  1120  kw.   and  the  line  voltage  and  current  are 
13,200  volts  and  49.8  amperes.     Oscillograph  records  show  that 
at  this  time  there  exists  across  diametrical  points  of  the  arma- 
ture  of  the   rotary  a  voltage  which  practically  consists  of  a 
fundamental  and  a  third  harmonic  that  is  8.5  per  cent,  of  the 
fundamental. 

Calculate  the  total  copper  loss  in  eath  transformer.  If  the 
low-tension  windings  had  been  divided  and  connected  in  double  Y 
with  no  connection  between  their  neutral  points  what  would  have 
been  the  copper  loss  in  each  transformer  under  the  same  load 
condition? 

44.  A3000-kw.,  5000-volt,  60-cycle  alternating-current  generator 
supplies  power  to  a  3-phase  transmission  line  through  three 
lOOO-kv.-a.  transformers  which  are  connected  in  Y  on  the  low- 
tension  and  in  delta  on  the  high-tension  sides.     With  the  low- 
tension  winding  short-circuited  and  with  3240  volts  at  60  cycles 
impressed  on  the  high-tension  side  full-load  current  exists  in  the 
windings  and  7490  watts  is  supplied.     The  ratio  of  transforma- 
tion is  2890  to  66,000  volts.     Both  the  neutral  point  of  the  arma- 
ture winding  and  the  neutral  point  of  the  low-tension  windings 
of  the  transformers  are  directly  connected  to  the  same  bus. 
When  the  transformers  are  delivering  2650  kw.  at  0.875  power 
factor  and  with  a  line  voltage  of  66,000  volts  they  become 


136  PROBLEMS  IN  ALTERNATING  CURRENT  MACHINERY 

unduly  hot.  Oscillograph  records  show  that  at  this  time  the 
terminal  voltage  of  each  phase  of  the  generator  practically  con- 
sists of  a  fundamental  and  a  third  harmonic  which  is  12  per 
cent,  of  the  fundamental. 

What  is  the  copper  loss  in  each  transformer?  If  the  neutral 
point  of  the  transformer  windings  is  disconnected  from  the 
ground  bus  what  will  be  the  copper  loss  in  each  transformer  for 
the  same  load  condition? 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
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WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  5O  CENTS  ON  THE  FOURTH 
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OVERDUE. 


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